Codeforces Round #394 (Div. 2) D. Dasha and Very Difficult Problem

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 sequencec = [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 sequencec = [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：

给你序列p和序列a，让你找一个合法的序列b，输出。

序列c是通过bi-ai来得到的。

序列p是通过离散化序列c来得到的。（离散化c序列，就是将c序列中的数字从小到大编号）

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
    int id,num;
}a[];
int b[],k[];
int n,l,r;
bool cmp(node a,node b)
{
    return a.id>b.id;
}
int main()
{
    scanf("%d%d%d",&n,&l,&r);
    for(int i=;i<=n;i++) scanf("%d",&a[i].num);
    for(int i=;i<=n;i++) {scanf("%d",&a[i].id); k[i]=a[i].id;}
    sort(a+,a+n+,cmp);
    b[]=r;
    bool flag=;
    for(int i=;i<=n;i++)
    {
        /*
        if (b[i-1]-a[i-1].num-1+a[i].num<l ||
                b[i-1]-a[i-1].num-1+a[i].num>r)
        {
            flag=0;
            break;
        }
       b[i]=b[i-1]-a[i-1].num-1+a[i].num;
       //以上是错误的,只考虑了差是连续，若是不连续没考虑。
       */
       if (b[i-]-a[i-].num-+a[i].num>=l)
       {
           b[i]=b[i-]-a[i-].num-+a[i].num;
           if (b[i]>r) b[i]=r;  //使差减小得更多。
       }
       else  { flag=; break; }
    }
    if (!flag) printf("-1");
       else{
             for(int i=;i<=n;i++)
             {
                 if (i-) printf(" ");
                 printf("%d",b[n-k[i]+]);
             }
           }
     printf("\n");
    return ;
}