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.
#ifndef __APPLE__
#include <bits/stdc++.h>
#else
#include <iostream>
#include <cstring>
#include <algorithm>
#endif

using namespace std;

typedef long long ll;
const int N = 105;
const int M = 1e9 + 7;
int n, c[N], b[N], q, x, bp[N], bpp[N];
ll dp[N][N * N], sdp[N][N * N], cp[N];

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> c[i];
        cp[i] = cp[i - 1] + c[i];
    }

    for (int i = 1; i <= n - 1; i++)
    {
        cin >> b[i];
        bp[i] = bp[i - 1] + b[i];
        bpp[i] = bpp[i - 1] + bp[i];
    }
    cin >> q;
    while (q--)
    {
        cin >> x;
        memset(dp, 0, sizeof(dp));
        memset(sdp, 0, sizeof(sdp));
        for (int i = 0; i < N; i++) sdp[0][i] = 1;
        for (int i = 1; i <= n; i++)
        {
            int minj = max(0, i * x + bpp[i - 1]);
            for (int j = minj; j <= cp[i]; j++)
            {
                sdp[i][j] = sdp[i][j] + sdp[i-1][min((int)cp[i-1], j)];
                if (j > c[i])
                    sdp[i][j] = (sdp[i][j] - sdp[i-1][j-c[i]-1]+M) % M;
            }
            for (int j = 1; j <= cp[i]; j++) {
                sdp[i][j] += sdp[i][j-1] + M;
                sdp[i][j] %= M;
            }
        }

        cout << sdp[n][cp[n]] << endl;
    }
    return 0;
}