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.
#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif //_MSC_VER

#include <cstdio>

#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <numeric>
#include <ctime>
#include <chrono>
#include <iostream>
#include <string>
#include <sstream>
#include <array>
#include <random>
#include <cmath>
#include <cassert>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <utility>
#include <thread>

using std::cin;
using std::cout;
using std::string;
using std::vector;

using ll = long long;

constexpr int MOD = 1000000000 + 7;  
 
vector<int> solve(vector<int> & c, vector<int> & b, vector<int> & x)
{
	int const n = c.size();
	int const q = x.size();
	vector<int> cb(n);
	for (int i = 1; i <= n - 1; ++i)
		cb[i] = cb[i - 1] + b[i - 1];
	vector<int> d(n);
	for (int i = 1; i <= n - 1; ++i)
		d[i] = d[i - 1] + cb[i];
	vector<int> cc(n + 1);
	for (int i = 1; i <= n; ++i)
		cc[i] = cc[i - 1] + c[i - 1];

	int const maximum = cc[n];
	vector<int> ans(q);
	for (int iq = 0; iq < q; ++iq)
	{
		vector< vector<int >>dp(n + 1, vector<int>(maximum + 1, 0));
		std::fill(dp[n].begin(), dp[n].end(), 1);

		for (int i = n - 1; i >= 0; --i)
		{
			for (int j = 0; j <= cc[i]; ++j)
			{
				for (int k = std::max(0, d[i] + (i + 1)*x[iq] - j); k <= c[i]; ++k)
				{
					assert(j + k <= maximum);
					dp[i][j] += dp[i + 1][j + k];
					dp[i][j] -= (dp[i][j] >= MOD ? MOD : 0);
				}
			}
		}

		ans[iq] = dp[0][0];
	}

	return ans;
}
static int solve(int testcase)
{
	vector<int> c, b, x;
	int n,q;
	cin >> n;
	c.assign(n, 0);
	for (int i = 0; i < n; ++i)
		cin >> c[i];

	b.assign(n - 1, 0);
	for (int i = 0; i < n - 1; ++i)
		cin >> b[i];

	cin >> q;
	x.resize(q);
	for (int i = 0; i < q; ++i)
		cin >> x[i];

	vector<int> ans = solve(c, b, x);
	for (int i = 0; i < q; ++i) {
		if (i)cout << ' ';
		cout << ans[i];
	}
	cout << std::endl;
	return 0;
}

int main(int argc, char* argv[])
{
 
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif //ONLINE_JUDGE
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);
	
	int tc = 1;
	//std::cin >> tc;
	for (int ic = 1; ic <= tc; ++ic)
	{
		solve(ic);
	}

	return 0;
}
