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>
#define maxn 105
#define maxN 10086

using namespace std;

const int p = 1e9 + 7;

int n, q, x;
int b[maxn], c[maxn], f[maxN], g[maxN], sb[maxn];

int cal(int x){
	for(int i = 0;i < maxN;i++) g[i] = 1;
	for(int i = 1;i <= n;i++){
		for(int j = 0;j < maxN;j++) f[j] = 0;
		for(int j = max(0, i * x + sb[i]);j < maxN;j++) f[j] = (g[j] + p - (j - c[i] - 1 >= 0 ? g[j - c[i] - 1] : 0)) % p;
		for(int j = 0;j < maxN;j++) g[j] = (f[j] + (j ? g[j - 1] : 0)) % p, f[j] = 0;
	}
	return g[maxN - 1];
}

map<int, int> mp;

int main(){
	scanf("%d", &n);
	int sum = 1;
	for(int i = 1;i <= n;i++) scanf("%d", &c[i]), sum = 1ll * sum * (c[i] + 1) % p;
	for(int i = 1;i < n;i++) scanf("%d", &b[i]);
	for(int i = 2;i <= n;i++) for(int j = 1;j < i;j++) sb[i] += (i - j) * b[j];
	int l = -((sb[n] + n - 1) / n), r = (n * maxn - sb[n] + n - 1) / n;
	for(int i = l;i <= r;i++) mp[i] = cal(i);
	scanf("%d", &q);
	while(q--){
		scanf("%d", &x);
		if(x < l) printf("%d\n", sum);
		else if(x > r) puts("0");
		else printf("%d\n", mp[x]);
	}
}