Problem: 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.

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
using namespace std;
#define endl '\n'
#define ll long long
#define pi pair<int, int>
#define f first
#define s second

const int mod = 1000000007;
const int m = 101, mxk = m * m;
int n, k, q;
int a[m], b[m], f[m], dp[mxk];

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	k = n * m;
	
	for(int i = 1; i <= n; i++) cin >> a[i];
	for(int i = 2; i <= n; i++) cin >> b[i], b[i] += b[i - 1];
	
	int s = 0x3f3f3f3f, e = 0x3f3f3f3f;
	for(int i = 1, j = 0; i <= n; i++){
		j += a[i], b[i] += b[i - 1];
		s = min(s, -b[i] / i - 1);
		e = min(e, (j - b[i]) / i + 1);
	} 
	
	for(int i = s; i <= e; i++){
		memset(dp, 0, sizeof(dp));
		dp[0] = 1;
		for(int j = 1; j <= n; j++){
			for(int l = 1; l <= k; l++) (dp[l] += dp[l - 1]) %= mod;
			for(int l = k; ~l; l--){
				(dp[l] += mod - (l < i * j + b[j] ? dp[l] : l > a[j] ? dp[l - a[j] - 1] : 0)) %= mod;
			}
		}
		for(int j = 0; j <= k; j++) (f[i - s] += dp[j]) %= mod;
	}
	
	cin >> q;
	
	while(q--){
		int x;
		cin >> x;
		x = max(s, min(e, x));
		cout << f[x - s] << endl;
	}
	
	return 0;
}