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 <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
inline int read()
{
	int f = 1, x = 0;
	char ch;

	do{
		ch = getchar();
		if (ch == '-')
			f = -1;
	}while(ch < '0' || ch > '9');
	do{
		x = x * 10 + ch - '0';
		ch = getchar();
	}while(ch >= '0' && ch <= '9');
	return f * x;
}
const int N = 100;
const int M = 2e5;
const int mod = 1e9 + 7;

int n, q;
int b[N + 1], c[N + 1], sumc[N + 1];
int f[N + 1][N * N + 1], g[N * N + 1], ans[M + 1];
int sum = 1, l = M / 2, r = M / 2;

inline int dp(int x)
{
	memset(f, 0, sizeof(int) * (N + 1));
	f[0][0] = 1;
	for (int i = 0; i <= N * N; i++)
		g[i] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = sumc[i]; j >= 0; j--)
			f[i][j] = (g[j] - ((j - c[i] > 0)?g[j - c[i] - 1]:0) + mod) % mod;
		for (int j = 0; j < x * i + b[i - 1]; j++)
			f[i][j] = 0;
		g[0] = f[i][0];
		for (int j = 1; j <= N * N; j++)
			g[j] = (g[j - 1] + f[i][j]) % mod;
	}
	return g[sumc[n]];
}
int main()
{
	n = read();
	for (int i = 1; i <= n; i++) {
		c[i] = read();
		sumc[i] = sumc[i - 1] + c[i];
		sum = 1ll * sum * (c[i] + 1) % mod;
	}
	for (int i = 1; i < n; i++)
		b[i] = b[i - 1] + read();
	for (int i = 1; i < n; i++)
		b[i] += b[i - 1];
	q = read();
	for (int i = 1; i <= n; i++) {
		l = min(l, (int)ceil(-b[i - 1] / i));
		r = min(r, (int)floor((sumc[i] - b[i - 1]) / i));
	}
	l = max(l, -M / 2);

	for (int i = l; i <= r; i++)
		ans[i + M / 2] = dp(i);

	for (int i = 1; i <= q; i++) {
		int x = read();
		printf("%d\n", x < l?sum:(x > r?0:ans[x + M / 2]));
	}
	return 0;
}
