Problem: This is the easy version of the problem. The only difference is that in this version q = 1. You can make hacks only if both versions of the problem are solved.

There is a process that takes place on arrays a and b of length n and length n-1 respectively. 

The process is an infinite sequence of operations. Each operation is as follows: 

  * First, choose a random integer i (1 ≤ i ≤ n-1). 
  * Then, simultaneously set a_i = min\left(a_i, \frac{a_i+a_{i+1}-b_i}{2}\right) and a_{i+1} = max\left(a_{i+1}, \frac{a_i+a_{i+1}+b_i}{2}\right) without any rounding (so values may become non-integer). 

See notes for an example of an operation.

It can be proven that array a converges, i. e. for each i there exists a limit a_i converges to. Let function F(a, b) return the value a_1 converges to after a process on a and b.

You are given array b, but not array a. However, you are given a third array c. Array a is good if it contains only integers and satisfies 0 ≤ a_i ≤ c_i for 1 ≤ i ≤ n.

Your task is to count the number of good arrays a where F(a, b) ≥ x for q values of x. Since the number of arrays can be very large, print it modulo 10^9+7.

Input

The first line contains a single integer n (2 ≤ n ≤ 100).

The second line contains n integers c_1, c_2 …, c_n (0 ≤ c_i ≤ 100).

The third line contains n-1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 100).

The fourth line contains a single integer q (q=1).

The fifth line contains q space separated integers x_1, x_2, …, x_q (-10^5 ≤ x_i ≤ 10^5).

Output

Output q integers, where the i-th integer is the answer to the i-th query, i. e. the number of good arrays a where F(a, b) ≥ x_i modulo 10^9+7.

Example

Input


3
2 3 4
2 1
1
-1


Output


56

Note

The following explanation assumes b = [2, 1] and c=[2, 3, 4] (as in the sample).

Examples of arrays a that are not good: 

  * a = [3, 2, 3] is not good because a_1 > c_1; 
  * a = [0, -1, 3] is not good because a_2 < 0. 



One possible good array a is [0, 2, 4]. We can show that no operation has any effect on this array, so F(a, b) = a_1 = 0.

Another possible good array a is [0, 1, 4]. In a single operation with i = 1, we set a_1 = min((0+1-2)/(2), 0) and a_2 = max((0+1+2)/(2), 1). So, after a single operation with i = 1, a becomes equal to [-1/2, 3/2, 4]. We can show that no operation has any effect on this array, so F(a, b) = -1/2.

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<bits/stdc++.h>
using namespace std;
#define N 105
#define ll long long
#define in(l,n)  for(int i=l;i<=n;i++)
int n,q,a[N],b[N],c[N],x[N];
ll sumb[N];
ll dp[N][N*N];
const int mod=1e9+7;
int main(){
    scanf("%d",&n);
    in(1,n)   scanf("%d",c+i);
    in(2,n)  {
        scanf("%d",b+i);
        b[i]=b[i]+b[i-1];
        sumb[i]=sumb[i-1]+b[i];
    }
    scanf("%d",&q);
    in(1,q)   scanf("%d",x+i);
    int M=*max_element(c+1,c+n+1);
    in(0,n*M) dp[0][i]=1;
    in(1,n){
        for(int j=0;j<=n*M;j++){
            if(j<i*x[1]+sumb[i])
                continue;
            dp[i][j]=dp[i-1][j];
            if(j>c[i])
                dp[i][j]-=dp[i-1][j-c[i]-1];
            dp[i][j]+=mod;
            dp[i][j]%=mod;
        }
        for(int j=1;j<=n*M;j++){
            dp[i][j]+=dp[i][j-1];
            dp[i][j]%=mod;
        }
    }
    dp[n][n*M]+=mod;
    dp[n][n*M]%=mod;
    printf("%lld\n",dp[n][n*M]);
}