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 <cstdio>
#include <map>
#include <algorithm>
#define int long long
const int nn = 105, mod = 1e9 + 7;
int n, q, f[nn * nn], g[nn * nn], b[nn], sb[nn], c[nn];
std::map<int, int> ans;
int solve(int x) {
	for (int i = 0; i < nn * nn; i++) g[i] = 1;
	for (int i = 1, lim; i <= n; i++) {
		lim = i * x + sb[i];
		for (int j = 0; j < nn * nn; j++) f[j] = 0;
		for (int j = std::max(0ll, lim); j < nn * nn; j++)
			if (j >= c[i] + 1) f[j] = (g[j] - g[j - c[i] - 1] + mod) % mod;
			else f[j] = g[j];
		g[0] = f[0];
		for (int j = 1; j < nn * nn; j++) g[j] = (g[j - 1] + f[j]) % mod;
//		for (int j = 0; j < nn * nn; j++) printf("%lld ", g[j]);
	}
//	puts("");
	return g[nn * nn - 1];
}
signed main() {
	scanf("%lld", &n);
	int prod = 1;
	for (int i = 1; i <= n; i++) {
		scanf("%lld", c + i);
		prod = prod * (c[i] + 1ll) % mod;
	}
	for (int i = 1; i < n; i++) scanf("%lld", b + i);
	for (int i = 1; i <= n; i++)
		for(int j = 1; j < i; j++)
			sb[i] += (i - j) * b[j];
	int lb = 0, rb = 2e9, m = *std::max_element(c + 1, c + n + 1);
	for (int i = 1; i <= n; i++) {
		lb = std::min(lb, -((sb[i] - 1) / i + 1));
		rb = std::min(rb, (n * m - sb[i] - 1) / i + 1);
	}
	for (int i = lb; i <= rb; i++) ans[i] = solve(i);
	scanf("%lld", &q);
	while (q--) {
		int x;
		scanf("%lld", &x);
		if (x < lb) printf("%lld\n", prod);
		if (x >= lb && x <= rb) printf("%lld\n", ans[x]);
		if (x > rb) puts("0");
	}
	return 0;
}