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 <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <array>
#include <random>
#include <chrono>
#include <string>
#include <queue>
#include <cassert>
#define int long long
#define pb push_back
#define pp pair<int,int>
#define ff first
#define dd second
#define mp make_pair

using namespace std;

int n, k, q, x, ans = 0;

int c[101];
int b[101];
int bsum[102];
const int maxSum = 100 * 100 + 2;
int mod = 1e9 + 7;
int dp[101][maxSum];
int psum[maxSum];

int getSum(int i) {
    return bsum[i+1];
}

void solve() {
    cin >> n;   
    for (int i = 0; i < n; ++i) cin >> c[i];
    bsum[0] = 0;
    for (int i = 0; i < n-1; ++i) { 
        cin >> b[i];
        bsum[i+1] = bsum[i] + b[i];
    }
    cin >> q;
    while (q--) {
        cin >> x;
        ans = 0;
        psum[0] = 0;
        for (int i = 0; i <= maxSum; ++i) {
            dp[0][i] = (int)(i >= x && i <= c[0]);
            psum[i+1] = psum[i] + dp[0][i];
        }
        int bb = 0;
        for (int i = 1; i < n; ++i) {
            bb += getSum(i-1);
            for (int sum = 0; sum < 100*n+1; ++sum) {
                dp[i][sum] = 0;
                if ((sum - bb) >= x*(i + 1)) {
                    dp[i][sum] += (psum[sum+1] - psum[max(sum-c[i], 0ll)] + mod) % mod; // dp[i-1][sum] + ... + dp[i-1][sum-c[i]];
                    dp[i][sum] %= mod;
                    if (i == n-1) {
                        ans += dp[i][sum];
                        ans %= mod;
                    }
                }
                //cout << "i sum dp[i][sum] " << i << " " << sum << " " << dp[i][sum] << endl;
            }
            for (int sum = 0; sum <= maxSum; ++sum) {
                psum[sum+1] = psum[sum] + dp[i][sum];
                psum[sum+1] %= mod;
            }
        }
        cout << ans << endl;
    }    
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    int z = 1;//; cin >> z;
    while (z--) {
        solve();
    }
}