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 <bits/stdc++.h>
const int N = 105, M = 3e6 + 10, offset = 1.5e6 + 5, P = 1e9 + 7;
int n, c[N], b[N], f[M], g[M], Q, x;
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &c[i]);
	for (int i = 1; i < n; i++)
		scanf("%d", &b[i]);
	scanf("%d%d", &Q, &x);
	int mx = c[1], mn = x;
	if (mx < 0) { puts("0"); return 0; }
	for (int i = std::max(0, mn); i <= mx; i++)
		f[i + offset] = 1;
	// for (int j = mn; j <= mx; j++)
		// printf("[%d]%d ", j, f[j + offset]);
	// puts("");
	int sum = 0;
	for (int i = 1; i < n; i++) {
		sum += b[i];
		// delta = a - sum \in [i * b[i], c[i + 1] + i * b[i]]
		// f[i + delta] <- f[i]
		// f[j] <- f[j - delta] j - delta \in [j - i * b[i] - c[i + 1], j - i * b[i]]
		memset(g, 0, sizeof g);
		mx += c[i + 1] + i * b[i]; mn = std::max(-(int)1e6, (i + 1) * (x + sum));
		for (int j = mn - i * b[i] - c[i + 1] - 1; j <= mx - i * b[i]; j++)
			g[j + offset] = (g[j - 1 + offset] + f[j + offset]) % P;
		memset(f, 0, sizeof f);
		for (int j = mn; j <= mx; j++)
			f[j + offset] = (g[j - i * b[i] + offset] - g[j - i * b[i] - c[i + 1] - 1 + offset] + P) % P;
		// for (int j = mn; j <= mx; j++)
			// printf("[%d]%d ", j, f[j + offset]);
		// puts("");
	}
	int ans = 0;
	for (int i = n * (x + sum); i <= 1.5e6; i++)
		ans = (ans + f[i + offset]) % P;
	printf("%d\n", ans);
	return 0;
}