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>
using namespace std;
using ll = long long;

const int N = 211;
const int mod = 1e9 + 7;

int n, c[N], b[N], q, x, d[N], s[N];
int dp[N][12315];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &c[i]);
	for (int i = 1; i < n; ++i)
		scanf("%d", &b[i]);
	scanf("%d%d", &q, &x);
	s[1] = x, d[1] = x;
	for (int i = 2; i <= n; ++i) {
		d[i] = d[i - 1] + b[i - 1];
		s[i] = s[i - 1] + d[i];
	}

	dp[0][0] = 1;
	int sc = 0;
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j <= sc; ++j)
			for (int k = 0; k <= c[i + 1]; ++k)
				if (j + k >= s[i + 1]) {
					dp[i + 1][j + k] = (dp[i + 1][j + k] + dp[i][j]) % mod;
				}
		sc += c[i + 1];
	}
	int ans = 0;
	for (int j = 0; j <= sc; ++j)
		ans = (ans + dp[n][j]) % mod;
	printf("%d\n", ans);
	return 0;
}

//~ int main() {
	//~ long long a, b, c, d;
	//~ cin >> a >> b  >> c >> d;
	//~ cout << a * 4 + b * 12 + c * 12 + d * 4 << endl;
	//~ long long a, b;
	//~ scanf("%lld%lld", &a, &b);
	//~ printf("%lld\n", __gcd(a, b));
	//~ printf("%lld %lld\n", a / __gcd(a,b), b / __gcd(a, b));
	//~ return 0;
//~ }

//~ C(52,10) = 15820024220
//~ C(48,9) = 1677106640
//~ C(48,8) = 377348994
//~ C(48,7) = 73629072
//~ C(48,6) = 12271512
//~ 12169249400 15820024220
