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

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本文介绍了一道复杂问题的解决思路，重点在于如何通过贪心算法找到符合条件的序列b，通过对输入序列p和a的处理，确保输出序列满足特定条件。

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: c_i = b_i - a_i.

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 ≤ a_i ≤ r and l ≤ b_i ≤ r. About sequence c we know that all its elements are distinct.

$\text{[math]}$

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 p_i equals to the number of integers which are less than or equal to c_i 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, ..., a_n (l ≤ a_i ≤ r) — the elements of the sequence a.

The next line contains n distinct integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ 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 c_i = b_i - a_i) has compressed sequence equals to p = [3, 2, 1, 4].

题目大意：

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。

思路：

1、既然序列p是通过序列c离散化得到的，那么其反应的就是一个从小到大的一个特征。那么首先我们按照p从小到大排序。

2、考虑ci=bi-ai.此时已知ai，求的是bi，那么bi=ci+ai；而ci是一个不定数组，其值只有大小顺序，而无固定值，那么我们要求的bi，只要合法，就是可行序列。

那么接下来我们贪心这个bi：

①设定p==1的位子上，bi=l，ci=l-ai；

②紧接着，设定p==2的位子上，尽可能的希望ci大于p==1的位子上的ci，并且p==2的位子上的ci尽可能的接近p==1的位子上的ci.

③依次类推，对于放置bi==r都不能满足ci>ci-1的情况下，那么就是-1的情况。

过程可能说的过于简单，具体参考代码（T T词穷了，不造咋说清楚些了）

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int pos;
    int num,val;
}a[100500];
int b[100500];
int ans[100500];
int c[100500];
int cmp(node a,node b)
{
    return a.val<b.val;
}
int main()
{
    int n,l,r;
    while(~scanf("%d%d%d",&n,&l,&r))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].num);
            a[i].pos=i;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].val);
        }
        int flag=0;
        sort(a+1,a+1+n,cmp);
        for(int i=1;i<=n;i++)
        {
            if(i==1)
            b[i]=l,c[i]=b[i]-a[i].num;
            else
            {
                if(r-a[i].num>c[i-1])
                {
                    b[i]=(c[i-1]+1+a[i].num);
                    if(b[i]<l)b[i]=l;
                    if(b[i]>r)flag=1;
                    c[i]=b[i]-a[i].num;
                }
                else flag=1;
            }
            ans[a[i].pos]=b[i];
        }
        if(flag==1)printf("-1\n");
        else
        {
            for(int i=1;i<=n;i++)
            {
                printf("%d ",ans[i]);
            }
            printf("\n");
        }
    }
}