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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int mod = 1000000007;

int main() {
    int n; cin >> n;
    vector<int> cs;
    for (int i=0; i<n; i++) {
        int t; cin >>t;
        cs.push_back(t);
    }
    vector<int> bs;
    for (int i=0; i<n-1; i++) {
        int t; cin >> t;
        bs.push_back(t);
    }
    int q; cin >> q;
    int x; cin >> x;

    vector<int> psums = vector<int>(n,0);
    for (int i=0; i<n; i++) {
        psums[i] = (i+1)*x;
        for (int j = i; j>0; j--) {
            psums[i] = psums[i] + j*bs[i-j];
        }
    }

    //dp[i][j] counts the number of a1,dots,ai satisfying the bounds
    //such that a1+dots+an = j
    vector<vector<int>> dp = vector<vector<int>>(n+1,vector<int>(10001,0));
    dp[0][0] = 1;
    int high=0;
    for (int i=1; i<=n; i++) {
        int low = max(0,psums[i-1]);
        high = high+cs[i-1];
        for (int j=low; j<=high; j++) {
            for (int k=0; k<= cs[i-1]; k++) {
                if (j-k>=0)
                    dp[i][j] = (dp[i][j]+dp[i-1][j-k])%mod;
            }
        }
    }

    int low = max(0,psums[n-1]);
    int ans = 0;
    for (int j=low; j<=high; j++) {
        ans = (ans+dp[n][j])%mod;
    }
    cout << ans << '\n';
}