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.
#include <stdio.h>
#include <iostream>
#include <string>
#include <string.h>
#include <vector>
#include <cmath>
#include <assert.h>
#include <algorithm>
#include <queue>
#include <climits>
#include <set>
#include <unordered_map>
#include <time.h>
#include <map>
#include <stack>
#include <unordered_set>
#include <bitset>
#define hash hassh
using namespace std;
int c[105],b[105];
long long dp[105][10005];
int preb[105];
int main() {
    int T = 1;
    //cin >> T;
    while (T--) {
        int n,q,x;
        cin>>n;
        for(int i=1;i<=n;i++) cin>>c[i];
        for(int i=1;i<n;i++) cin>>b[i];
        cin>>q>>x;
        for(int i=2;i<=n;i++){
            preb[i]=preb[i-1];
            for(int j=1;j<i;j++) preb[i]+=b[j];
        }
        long long ans=0,mod=1000000007;
        dp[0][0]=1;
        for(int i=1;i<=n;i++){
            int mn=i*x+preb[i];
            for(int j=max(mn,0);j<=i*100;j++){
                for(int k=0;k<=c[i];k++)
                    if(j>=k)
                        dp[i][j]+=dp[i-1][j-k],dp[i][j]%=mod;
                if(i==n) ans+=dp[i][j],ans%=mod;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}
//注意检查边界
//注意考虑某些数组/向量可能为空 可能引起错误
//注意各种边界是0还是1
//注意少加不必要的限制条件 避免出错
//不要把前期题想得太难，尝试找简单结论
//想想直接套路化dp行不行！
//求倍增要一层一层求不要一项一项求！！！！
