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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define int ll
#define ld long double
#define pii pair<int, int>
#define f first
#define s second
#define boost() cin.tie(0), cin.sync_with_stdio(0)

const int MN = 105, MOD = 1e9 + 7;

int n, a[MN], b[MN], c[MN], psa[MN], psum[MN], q, x, dp[12000005], ndp[12000005];

int32_t main() {
    boost();

    cin >> n;
    for (int i = 1; i <= n; i++) cin >> c[i], psa[i] = psa[i - 1] + c[i];
    for (int i = 1; i < n; i++) cin >> b[i];
    cin >> q >> x;
    int sum = 0;
    a[1] = x;
    for (int i = 2; i <= n; i++) a[i] = a[i - 1] + b[i - 1];
    for (int i = 1; i <= n; i++) sum += a[i], psum[i] = psum[i - 1] + a[i];
    //printf("%lld\n", sum);
    //if (x > c[1]) return 0 * printf("0\n");
    int ans = 0;
    dp[0] = 1;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= psa[i - 1]; j++) {
            int lo = 0, hi = c[i];
            //if (i == 1) lo = max(lo, x);
            ndp[j + lo] += dp[j];
            ndp[j + lo] %= MOD;
            ndp[j + hi + 1] -= dp[j];
            ndp[j + hi + 1] %= MOD;
        }
        for (int j = 1; j <= psa[i]; j++) {
            ndp[j] += ndp[j - 1];
            ndp[j] %= MOD;
        }
        for (int j = 0; j <= psa[i]; j++) {
            dp[j] = ndp[j];
            if (j < psum[i]) dp[j] = 0;
        }
        //printf("\n");
        for (int j = 0; j <= psa[i] + 1; j++) {
            ndp[j] = 0;
        }
    }
    for (int i = 0; i <= psa[n]; i++) ans += dp[i], ans %= MOD;
    ans %= MOD, ans += MOD, ans %= MOD;
    printf("%lld\n", ans);

    return 0;
}