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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define y1 y114514
#define pb push_back
#define mkp make_pair
#define fi first
#define se second
#define all(a) a.begin(), a.end()
const int M = 1000000007;

const int maxn = 105;
int n;
int c[maxn];
int b[maxn];
int sum[maxn];
int ss[maxn];
int tot[maxn];
int dp[maxn][maxn * 100];
int q, x;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", c + i), tot[i] = tot[i - 1] + c[i];
    for (int i = 1; i < n; ++i) {
        scanf("%d", b + i);
        sum[i + 1] = sum[i] + b[i];
        ss[i + 1] = ss[i] + sum[i + 1];
    }
    scanf("%d%d", &q, &x);
    dp[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j <= tot[i]; ++j)
            if (dp[i - 1][j]) {
                int limit = x * i + ss[i] - j;
                // printf("i:%d j:%d limit:%d\n", i, j, limit);
                for (int k = max(0, limit); k <= c[i]; ++k) {
                    dp[i][j + k] = dp[i][j + k] + dp[i - 1][j];
                    if (dp[i][j + k] >= M) dp[i][j + k] -= M;
                    // printf("dp[%d][%d]->%d->dp[%d][%d], cur: %d\n", i - 1, j, k, i, j + k, dp[i][j + k]);
                }
            }
    }

    // for (int i = 1; i <= n; ++i)
    //     for (int j = 0; j <= tot[i]; ++j) printf("dp[%d][%d]: %d\n", i, j, dp[i][j]);

    int ans = 0;
    for (int i = 0; i <= tot[n]; ++i) (ans += dp[n][i]) %= M;
    printf("%d\n", ans);
    return 0;
}