CodeForces - 761D Dasha and Very Difficult Problem （構造）

題目鏈接；http://codeforces.com/problemset/problem/761/D點擊打開鏈接

D. Dasha and Very Difficult Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Dasha logged into the system and began to solve problems. One of them is as follows:

Given two sequences a and b of length n each you need to write a sequence c of length n, the i-th element of which is calculated as follows: ci = bi - ai.

About sequences a and b we know that their elements are in the range from l to r. More formally, elements satisfy the following conditions: l ≤ ai ≤ r and l ≤ bi ≤ r. About sequence c we know that all its elements are distinct.

Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence a and the compressed sequence of the sequence c were known from that test.

Let's give the definition to a compressed sequence. A compressed sequence of sequence c of length n is a sequence p of length n, so that pi equals to the number of integers which are less than or equal to ci in the sequence c. For example, for the sequence c = [250, 200, 300, 100, 50] the compressed sequence will be p = [4, 3, 5, 2, 1]. Pay attention that in c all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to n inclusively.

Help Dasha to find any sequence b for which the calculated compressed sequence of sequence c is correct.

Input

The first line contains three integers n, l, r (1 ≤ n ≤ 105, 1 ≤ l ≤ r ≤ 109) — the length of the sequence and boundaries of the segment where the elements of sequences a and b are.

The next line contains n integers a1,  a2,  ...,  an (l ≤ ai ≤ r) — the elements of the sequence a.

The next line contains n distinct integers p1,  p2,  ...,  pn (1 ≤ pi ≤ n) — the compressed sequence of the sequence c.

Output

If there is no the suitable sequence b, then in the only line print "-1".

Otherwise, in the only line print n integers — the elements of any suitable sequence b.

Examples

input

5 1 5
1 1 1 1 1
3 1 5 4 2

output

3 1 5 4 2 

input

4 2 9
3 4 8 9
3 2 1 4

output

2 2 2 9 

input

6 1 5
1 1 1 1 1 1
2 3 5 4 1 6

output

-1

Note

Sequence b which was found in the second sample is suitable, because calculated sequence c = [2 - 3, 2 - 4, 2 - 8, 9 - 9] = [ - 1,  - 2,  - 6, 0] (note that ci = bi - ai) has compressed sequence equals to p = [3, 2, 1, 4].

ci=bi-ai

pi爲ci在c數組中爲第pi小個

給你p數組和a數組 構造b數組且範圍不可超過l~r

我們從最小的ci開始 爲了使後面更有空間構造 從範圍的左端點開始構造b數組 當碰見某個b小於範圍l時 將最小值加至l 

如果超過r則說明無法構造 因爲此時構造已爲c數組的最小情況

#include <bits/stdc++.h>
using namespace std;
int n,l,r;
struct xjy
{
    int a;
    int num;
    int pos;
    int b;
    bool operator < (const xjy &r)const
    {
        return num<r.num;
    }
};
bool cmp1(xjy a1,xjy a2)
{
    return a1.pos<a2.pos;
}
vector<xjy > s;
int main()
{
    cin >> n >>l >>r;
    for(int i=0;i<n;i++)
    {
        xjy mid;
        cin >> mid.a;
        mid.pos=i;
        s.push_back(mid);
    }
    for(int i=0;i<n;i++)
        cin >> s[i].num;
    sort(s.begin(),s.end());
    int mid=0;
    int cnt=l-s[0].a;
    s[0].b=cnt+s[0].a;
    cnt++;
    for(int i=1;i<n;i++)
    {
        s[i].b=cnt+s[i].a;
        cnt++;
        if(s[i].b>r)
        {
            cout << -1 ;
            return 0;
        }
        if(s[i].b<l)
        {
            mid=max(mid,l-s[i].b);
        }
    }
    sort(s.begin(),s.end(),cmp1);
    for(int i=0;i<n;i++)
    {
        if(s[i].b+mid>r)
        {
            cout << "-1" <<endl;
            return 0;
        }
    }
    for(int i=0;i<n;i++)
        cout <<s[i].b+mid << " ";
}