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

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 weightai.

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:

the first monster can't eat the second monster because a1 = 1 is not greater than a2 = 2;
the second monster can't eat the third monster because a2 = 2 is not greater than a3 = 2;
the second monster can't eat the fifth monster because they are not neighbors;
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, thej-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 thex-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：給出一個初始序列和一個目標序列，讓你對初始序列進行一些合併操作，使得初始序列變成目標序列。

每次操作可以在初始序列中選一個數，然後讓它和相鄰的一個比它小的數合併，合併後的值是這兩個數的和。

分析：很容易想到如果初始序列可以轉換成目標序列，那麼可以根據目標序列將原序列分割，而且這個分割是固定的。如果初始序列有合理的分割方案而且存在合理的合併方案，那麼必定能轉換成目標序列。（每一段的合併的起點可以設爲該段的最大值，當最大值不止一個時，要選一個可以進行合併的最大值作爲起點）

#include <bits/stdc++.h>
using namespace std;

struct node
{
    int l, r, m;
    node() {}
    node(int _l, int _r, int _m):l(_l),r(_r),m(_m) { }
};

int a[600], b[600];
vector<node>v;

int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
    }
    
    int m;
    scanf("%d", &m);
    for(int j = 1; j <= m; j++)
    {
        scanf("%d", &b[j]);
    }
    
    int cnt = 0;
    int flag = 1, i, j, k = 1, index = 0;
    for(i = 1, j = 1; i <= n; i++)
    {
        cnt += a[i];
        if(j == m + 1) 
        {
            flag = 0;
            break;
        }
        if(a[i] > a[index]) index = i;
        if(cnt == b[j])
        {
            if(index != k || i ==k);
            else 
            {
                int tr = index + 1;
                while(tr <= i && a[index] == a[tr]) tr++;
                if(tr > i) 
                {
                    flag = 0;
                    break;
                }
                index = tr - 1;
            } // index爲每一段進行合併的起點
            v.push_back(node(k, i, index));
            index = 0;
            cnt = 0;
            j++;
            k = i + 1; 
        }
        else if(cnt > b[j])
        {
            flag = 0;
            break;
        }
    }
    if(j != m + 1)
        flag = 0;
    
    if(!flag) puts("NO");
    else
    {
        puts("YES");
        int cnt = 0;
        for(int i = 0; i < v.size(); i++)
        {
            if(a[v[i].m] > a[v[i].m + 1])
            {
                int tmp = v[i].m - cnt;
                for(int j = v[i].m; j < v[i].r; j++)
                {
                    printf("%d R\n", tmp);
                    cnt++;
                }
                int tp=tmp;
                for(int j = v[i].m; j > v[i].l; j--)
                {
                    printf("%d L\n", tp);
                    cnt++;
                    tp--;
                }
            }
            else
            {
                for(int j = v[i].m; j > v[i].l; j--)
                {
                    printf("%d L\n", v[i].m - cnt);
                    cnt++;
                }
                int tmp = v[i].m - cnt;
                for(int j = v[i].m; j < v[i].r; j++)
                {
                    printf("%d R\n", tmp);
                    cnt++;
                }
            }
        }
    }
    
    return 0;
}