Codeforces Round #378 (Div. 2) C.Epidemic in Monstropolis (分塊處理）

C. Epidemic in Monstropolis

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There was an epidemic in Monstropolis and all monsters became sick. To recover, all monsters lined up in queue for an appointment to the only doctor in the city.

Soon, monsters became hungry and began to eat each other. 

One monster can eat other monster if its weight is strictly greater than the weight of the monster being eaten, and they stand in the queue next to each other. Monsters eat each other instantly. There are no monsters which are being eaten at the same moment. After the monster A eats the monster B, the weight of the monster A increases by the weight of the eaten monster B. In result of such eating the length of the queue decreases by one, all monsters after the eaten one step forward so that there is no empty places in the queue again. A monster can eat several monsters one after another. Initially there were n monsters in the queue, the i-th of which had weight ai.

For example, if weights are [1, 2, 2, 2, 1, 2] (in order of queue, monsters are numbered from 1 to 6 from left to right) then some of the options are:

1. the first monster can't eat the second monster because a1 = 1 is not greater than a2 = 2; 
2. the second monster can't eat the third monster because a2 = 2 is not greater than a3 = 2; 
3. the second monster can't eat the fifth monster because they are not neighbors; 
4. the second monster can eat the first monster, the queue will be transformed to [3, 2, 2, 1, 2]. 

After some time, someone said a good joke and all monsters recovered. At that moment there were k (k ≤ n) monsters in the queue, the j-th of which had weight bj. Both sequences (a and b) contain the weights of the monsters in the order from the first to the last.

You are required to provide one of the possible orders of eating monsters which led to the current queue, or to determine that this could not happen. Assume that the doctor didn't make any appointments while monsters were eating each other.

Input

The first line contains single integer n (1 ≤ n ≤ 500) — the number of monsters in the initial queue.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the initial weights of the monsters.

The third line contains single integer k (1 ≤ k ≤ n) — the number of monsters in the queue after the joke. 

The fourth line contains k integers b1, b2, ..., bk (1 ≤ bj ≤ 5·108) — the weights of the monsters after the joke. 

Monsters are listed in the order from the beginning of the queue to the end.

Output

In case if no actions could lead to the final queue, print "NO" (without quotes) in the only line. 

Otherwise print "YES" (without quotes) in the first line. In the next n - k lines print actions in the chronological order. In each line print x — the index number of the monster in the current queue which eats and, separated by space, the symbol 'L' if the monster which stays the x-th in the queue eats the monster in front of him, or 'R' if the monster which stays the x-th in the queue eats the monster behind him. After each eating the queue is enumerated again. 

When one monster eats another the queue decreases. If there are several answers, print any of them.

Examples

input

6
1 2 2 2 1 2
2
5 5

output

YES
2 L
1 R
4 L
3 L

input

5
1 2 3 4 5
1
15

output

YES
5 L
4 L
3 L
2 L

input

5
1 1 1 3 3
3
2 1 6

output

NO

Note

In the first example, initially there were n = 6 monsters, their weights are [1, 2, 2, 2, 1, 2] (in order of queue from the first monster to the last monster). The final queue should be [5, 5]. The following sequence of eatings leads to the final queue:

• the second monster eats the monster to the left (i.e. the first monster), queue becomes [3, 2, 2, 1, 2]; 
• the first monster (note, it was the second on the previous step) eats the monster to the right (i.e. the second monster), queue becomes [5, 2, 1, 2]; 
• the fourth monster eats the mosnter to the left (i.e. the third monster), queue becomes [5, 2, 3]; 
• the finally, the third monster eats the monster to the left (i.e. the second monster), queue becomes [5, 5]. 

Note that for each step the output contains numbers of the monsters in their current order in the queue.

題意：

給你n個數字排成一個隊列a，每個數字可以吃掉比自己兩邊相鄰的比自己小的數字，吃完一個數字後整個隊列位於該數字之後的會集體向前前進一個。之後給你k個數字組成一個隊列b，問你是否有可能讓n個數字進行多次相互吞吃之後形成這個隊列b。

分析：

當時剛做完第二題看第三題才這麼點人過，人數居然比第四題還少很吃驚，結果發現第四題的坑果然比第三題少多了。解法很簡單，你遍歷k個數字組成的隊列b，如果能通過數字吞吃形成這個隊列，那麼在隊列a中必定存在前i個數字相互吞吃能夠形成b1，i由你自己計算。換句話說，隊列b把a中n個數字分成了k份，每一份都能夠組成隊列b中的一個元素，從而形成k個元素的隊列b。

那麼要做的就很簡單了，開個循環遍歷b，在循環裏再遍歷a即可。做法很簡單但是敲起來倒不短，wa了不少次。

#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define INF 0x3f3f3f3f
const long long N=505;
const long long mod=1e9;
const double PI=acos(-1.0);
typedef long long ll;

int a[N],b[N];

int main() {
    int n,k;
    while (cin>>n) {
        long long x=0,y=0;
        for (int i=1; i<=n; i++) {
            scanf("%d",&a[i]);
            y+=a[i];
        }
        cin>>k;
        for (int i=1; i<=k; i++) {
            scanf("%d",&b[i]);
            x+=b[i];
        }
        if (x!=y) {
            cout<<"NO\n";
            continue;
        }
        int flag=0,cur=1,temp=0,sssr=0;
        for (int i=1; i<=k; i++) {
            temp=0;
            flag=cur;
            while (temp<b[i]) {
                temp+=a[cur];
                cur++;
                //                cout<<temp<<endl;
            }
            if (temp!=b[i]) {
                sssr=1;
                break;
            }
            int ssr=0;
            if (cur-flag==1) {
                ssr=1;
            }
            for (int j=flag+1; j<cur; j++) {
                if (a[j]!=a[flag]) {
                    ssr=1;
                    break;
                }
            }
            if (ssr==0) {
                sssr=1;
                break;
            }
        }
        if (sssr==1) {
            cout<<"NO\n";
            continue;
        } else {
            cout<<"YES\n";
            cur=1;
            int sum=0;
            for (int i=1; i<=k; i++) {
                temp=0;
                flag=cur;
                while (temp<b[i]) {
                    temp+=a[cur];
                    cur++;
                }
                
                int pos=flag,maxx=a[flag];
                for (int j=flag+1; j<cur; j++) {
                    if (a[j]>maxx) {
                        maxx=a[j];
                        pos=j;
                    }
                }
                int en=1;
                if (pos!=flag) {
                    for (int j=pos; j>flag; j--) {
                        en++;
                        printf("%d L\n",j-sum);
                    }
                    for (int j=0; j<cur-flag-en; j++) {
                        printf("%d R\n",flag-sum);
                    }
                } else {
                    for (int j=flag+1; j<cur-1; j++) {
                        if (a[j]>=maxx&&a[j]>a[j+1]) {
                            pos=j;
                        }
                    }
                    
                    for (int j=pos; j<cur-1; j++) {
                        printf("%d R\n",pos-sum);
                    }
                    en--;
                    for (int j=pos; j>flag; j--) {
                        printf("%d L\n",pos-en-sum);
                        en++;
                    }
                }
                sum+=cur-1-flag;
            }
        }
    }
    return 0;
}