Codeforces 761D-Dasha and Very Difficult Problem

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].

题意：数组a和数组b中每个元素都是不小于l，不大于r，数组c由数组b中元素减去数组a中对应位置的元素得到，即ci = bi - ai，数组p表示相应位置的数组c中的元素在数组c中排第几，现在告诉你l,r和数组a，数组p，若能得到数组b

，则任写一组数组b，否则输出-1

解题思路：用结构体的形式来保存数组a，数组b，数组p，数组c中的各个元素，先以数组p来排序，则第一个b元素应

尽量小，所以可以定为l，后面的元素可以以此来推，若推出来的b元素小于l，则可以将其变大至l，若大于r，则说明

无法找到这样的数组b

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

struct node
{
    int a,b,p,c;
    int id;
}x[100009];
int n,l,r;

bool cmp(node xx,node yy)
{
    return xx.p<yy.p;
}

bool cmp1(node xx,node yy)
{
    return xx.id<yy.id;
}

int main()
{
    while(~scanf("%d %d %d",&n,&l,&r))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x[i].a);
            x[i].id=i;
        }
        for(int i=1;i<=n;i++)
            scanf("%d",&x[i].p);
        sort(x+1,x+1+n,cmp);
        int flag=1;
        for(int i=1;i<=n;i++)
        {
            if(i==1) x[i].b=l,x[i].c=l-x[i].a;
            else
            {
                x[i].c=x[i-1].c+1;
                x[i].b=x[i].a+x[i].c;
                if(x[i].b<l) x[i].b=l,x[i].c=l-x[i].a;
                if(x[i].b>r) flag=0;
            }
            if(flag==0) break;
        }
        if(!flag) printf("-1\n");
        else
        {
            sort(x+1,x+1+n,cmp1);
            printf("%d",x[1].b);
            for(int i=2;i<=n;i++)
                printf(" %d",x[i].b);
            printf("\n");
        }
    }
    return 0;
}