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>
using namespace std;
#define ll long long
const int maxn = 105;
const int maxm = 10005;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
	x += y;
	return x >= mod ? x - mod : x;
}
inline int sub(int x, int y) {
	x -= y;
	return x < 0 ? x + mod : x;
}
inline int mul(int x, int y) {
	return 1ll * x * y % mod;
}
int n, q, c[maxn], b[maxn], preb[maxn];
int _ans[maxm << 1], *ans = _ans + maxm;
int calc(int x) {
	static int f[maxm], _s[maxm], *sum = _s + 1;
	f[0] = 1;
	int low = 0, upp = 0;
	for(int i = 1; i <= n; i++) {
		sum[low - 1] = 0;
		for(int j = low; j <= upp; j++)
			sum[j] = add(sum[j - 1], f[j]);
		int st = max(0, x * i + preb[i]), ed = upp + c[i];
		if(st > ed)
			return 0;
		for(int j = st; j <= ed; j++) 
			f[j] = sub(sum[min(j, upp)], sum[max(j - c[i], low) - 1]);
		low = st, upp = ed;
	}
	int ans = 0;
	for(int i = low; i <= upp; i++)
		ans = add(ans, f[i]);
	return ans;
}
int main() {
	scanf("%d", &n);
	int all = 1;
	for(int i = 1; i <= n; i++) 
		scanf("%d", &c[i]), all = mul(all, c[i] + 1);	
	for(int i = 1; i < n; i++) {
		scanf("%d", &b[i]);		
		preb[i + 1] = preb[i] + b[i];
	}
	for(int i = 2; i <= n; i++)
		preb[i] += preb[i - 1];
	int L = 0, R = 1;
	ans[0] = calc(0), ans[1] = calc(1);
	for(; ans[R]; R++, ans[R] = calc(R));
	for(; ans[L] != all; L--, ans[L] = calc(L));
	scanf("%d", &q);
	for(int i = 1, x; i <= q; i++) {
		scanf("%d", &x);
		if(x > R)
			puts("0");
		else if(x < L)
			printf("%d\n", all);
		else
			printf("%d\n", ans[x]);
	}
	return 0;
}