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<bits/stdc++.h>
using namespace std;

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

int c[N], b[N];
int bs[N], bp[N];
int dp[N][N * N];

int n, q;
int rlt[N];

int main() {
	
	cin >> n;
	int M = 0, res = 1;
	for (int i = 1; i <= n; i ++) {
		cin >> c[i];
		if (c[i] > M) M = c[i];
		res = 1LL * res * (c[i] + 1) % MOD;
	}
	for (int i = 2; i <= n; i ++) {
		cin >> b[i];
		bs[i] = bs[i - 1] + b[i];
	}
	for (int i = 2; i <= n; i ++) bp[i] = bp[i - 1] + bs[i];
	int LX = -bp[n] / n, RX = LX + M + 1;
	for (int x = LX; x <= RX; x ++) {
		for (int i = 0; i <= n; i ++) for (int j = 0; j <= M * n; j ++) dp[i][j] = 0;
		dp[0][0] = 1;
		for (int i = 1; i <= n; i ++) {
			int L = i * x + bp[i], sum = 0;
			for (int j = 0; j <= n * M; j ++) {
				if (j >= 1) dp[i][j] = dp[i][j - 1];
				(dp[i][j] += dp[i - 1][j]) %= MOD;
				if (j > c[i]) (dp[i][j] += MOD - dp[i - 1][j - c[i] - 1]) %= MOD;
			}
			for (int j = 0; j < L; j ++) dp[i][j] = 0;
		}
		rlt[x - LX] = 0;
		for (int i = 0; i <= n * M; i ++) (rlt[x - LX] += dp[n][i]) %= MOD;
	}
	cin >> q;
	while (q --) {
		int x; cin >> x;
		if (x < LX) cout << res << endl;
		else if (x > RX) cout << 0 << endl;
		else cout << rlt[x - LX] << endl;
	}
}