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 int long long

const int N = 105;
const int MOD = 1000000007;

int n;
int suff[N];
int dp[N][N * N];
int pref[N][N * N];
int val[N], c[N], b[N];
int cnt = 0;
int x[N];

void backtrack(int k) {
    for (int i = 0; i <= c[k]; i++) {
        x[k] = i;
        if (k == n) {
            int sum = 0;
            bool ok = 1;
            for (int i = 1; i <= n; i++) {
                sum += x[i];
                if (sum < val[i]) ok = 0;
            }
            cnt += ok;
        } else backtrack(k + 1);
    }
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> n;
    for (int i = 1; i <= n; i++) cin >> c[i];
    for (int i = 1; i < n; i++) cin >> b[i]; 
    int q, x;
    cin >> q >> x;
    int cur = x;
    for (int i = 1; i <= n; i++) {
        val[i] = val[i - 1] + cur;
        cur += b[i];
        // cout << val[i] << '\n';
    }
    // backtrack(1);
    suff[n + 1] = 1;
    for (int i = n; i >= 1; i--) suff[i] = suff[i + 1] * (c[i] + 1) % MOD;
    dp[0][0] = 1;
    for (int j = 0; j < N*N; j++) pref[0][j] = 1;
    int res = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < N * N; j++) {
            int l = max<int>(0, j - c[i]);
            int r = j;
            dp[i][j] = pref[i - 1][r] - (l > 0 ? pref[i - 1][l - 1] : 0);
            dp[i][j] = (dp[i][j] + MOD) % MOD;
        }
        // for (int j = 0; j < 10; j++) cout << dp[i][j] << ' ';
        // cout << endl;
        for (int j = 0; j < min(N * N, val[i]); j++) {
            res = (res + dp[i][j] * suff[i + 1]) % MOD;
            dp[i][j] = 0;
        }
        pref[i][0] = dp[i][0];
        for (int j = 1; j < N * N; j++) pref[i][j] = (dp[i][j] + pref[i][j - 1]) % MOD;
    }
    // cout << res << endl;
    res = (suff[1] - res + MOD) % MOD;
    cout << res << '\n';

    return 0;
}
