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 <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <fstream>
#include <random>
#include <iomanip>
#include <clocale>
#include <bitset>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>

using namespace std;

#define endl '\n'
#define int long long

const int MOD = 1000000007;
const int INF = 2000000000;
const int MAXN = 100000;

int dp[100][20000];

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<int> c(n);
    for (int i = 0; i < n; i++) {
        cin >> c[i];
    }
    vector<int> b(n);
    for (int i = 1; i < n; i++) {
        cin >> b[i];
    }
    vector<int> bp(n);
    for (int i = 1; i < n; i++) {
        bp[i] = bp[i - 1] + b[i];
    }
    for (int i = 1; i < n; i++) {
        bp[i] += bp[i - 1];
    }
    int q;
    cin >> q;
    for (int _ = 0; _ < q; _++) {
        int x;
        cin >> x;
        vector<int> need(n);
        for (int i = 0; i < n; i++) {
            need[i] = bp[i] + x * (i + 1);
        }
        for (int i = 0; i < 100; i++) {
            for (int j = 0; j <= 10000; j++) {
                dp[i][j] = 0;
            }
        }
        for (int i = max(0LL, need[0]); i <= c[0]; i++) {
            dp[0][i] = 1;
        }
        for (int i = 1; i < n; i++) {
            for (int j = need[i]; j <= 10000; j++) {
                for (int k = 0; k <= c[i]; k++) {
                    if (j < k) {
                        break;
                    }
                    dp[i][j] += dp[i - 1][j - k];
                    if (dp[i][j] >= MOD) {
                        dp[i][j] -= MOD;
                    }
                }
            }
        }
        int ans = 0;
        for (int j = 0; j <= 10000; j++) {
            ans += dp[n - 1][j];
        }
        cout << ans % MOD;
    }
}