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.
#pragma GCC optimize("O3","Ofast","no-stack-protector","unroll-loops","omit-frame-pointer","inline") //Optimization flags
#pragma GCC option("arch=native","tune=native","no-zero-upper") //Enable AVX
// #pragma GCC target("avx2") //Enable AVX

#include <bits/stdc++.h>

#define int long long

using namespace std;

const int MOD = 1e9 + 7;

const int MAXN = 110;

int C[MAXN];

int B[MAXN];

int T[MAXN];

int dp[MAXN][MAXN*MAXN];

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

	int N;
	cin >> N;

	for(int i = 1 ; i <= N ; i++)
	{
		cin >> C[i];
	}    

	for(int i = 2 ; i <= N ; i++)
	{
		cin >> B[i];
		T[i] = T[i-1]+B[i];
	}

	for(int i = 1 ; i <= N ; i++)
	{
		T[i] += T[i-1];
	}

	int ini = -T[N]/N - 50;
	int fin = (100*N-T[N])/N + 50;

	map<int,int> mp;

	for(int X = ini ; X <= fin ; X++)
	{
		for(int j = 0 ; j <= 10000 ; j++)
		{
			dp[0][j] = 1;
		}

		for(int i = 1 ; i <= N ; i++)
		{
			for(int j = 0 ; j <= 10000 ; j++)
			{
				dp[i][j] = 0;
				if(j-T[i] >= X*i)
				{
					int val = 0;
					if(j-C[i]-1 >= 0)val = dp[i-1][j-C[i]-1];

					dp[i][j] = (dp[i-1][j]+MOD-val);
					dp[i][j] %= MOD;
				}
			}

			for(int j = 1 ; j <= 10000 ; j++)
			{
				dp[i][j] += dp[i][j-1];
				dp[i][j] %= MOD;
			}
		}

		mp[X] = dp[N][10000];
	}

	int Q;
	cin >> Q;

	for(int i = 1 ; i <= Q ; i++)
	{
		int X;
		cin >> X;

		if(X < ini)cout << mp[ini] << '\n';
		else if(X > fin)cout << 0 << '\n';
		else cout << mp[X] << '\n';
	}

    return 0;
}