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<bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i < b; i ++)
#define LL long long
#define mx 105
using namespace std;

const LL mod = 1000000007;

int c[mx], b[mx], n, bb, B[mx];
int st, en;

LL res[300], dp[mx][mx * mx];
LL add[mx * mx];

void Add(LL &x, LL y){
	x = (x + y + mod) % mod;
}

void calc(int x){
	memset(dp, 0, sizeof dp);
	rep(i, max(x, 0), c[1] + 1) dp[1][i] = 1;
	int lim = x;
	rep(i, 2, n + 1) {
		lim += x + b[i - 1];
		memset(add, 0, sizeof add);
		rep(j, 0, 10001) if(dp[i - 1][j] && j + c[i] >= lim){
			Add(add[j + max(lim - j, 0)], dp[i - 1][j]);
			Add(add[j + c[i] + 1], -dp[i - 1][j]);
		}
		rep(j, 0, 10001){
			j ? dp[i][j] = (dp[i][j - 1] + add[j]) % mod : dp[i][j] = add[j];
		}
	}
	rep(j, 0, 10001) Add(res[x - st], dp[n][j]);
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
#endif
	
	cin >> n;
	rep(i, 1, n + 1) cin >> c[i];
	rep(i, 1, n) {
		cin >> bb;
		b[i] = b[i - 1] + bb;
		B[i] = B[i - 1] + b[i];
	}
	st = -B[n - 1] / n - 2;
	en = 100 + st + 5;
	rep(x, st, en + 1) calc(x);
	int q, X;
	cin >> q;
	while(q --){
		cin >> X;
		if(X < st) cout << res[0] << endl;
		else if(X <= en) cout << res[X - st] << endl;
		else cout << 0 << endl;
	}
	
	return 0;
}