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<iostream>
#include<algorithm>
using namespace std;

long long ans = 0, MM = 1000000007;
int n;
int b[110], c[110], q, x, pivot;
long long t[10010], old_t[10010];
long long t1[10010], old_t1[10010];
int main()
{
	ios_base::sync_with_stdio(false); cin.tie(NULL);
	int i, j, k;
	cin >> n;
	for (i = 1; i <= n; i++) cin >> c[i];
	pivot = 0;
	for (i = 1; i < n; i++) cin >> b[i];
	cin >> q;
	cin >> x;

	for (j = 0; j <= 10000; j++) old_t[j] = 0;
	old_t[0] = old_t1[0] = 1;
	for (i = 1; i <= n; i++)
	{
		pivot = i * x;
		for (j = 1; j < i; j++) pivot += (i - j) * b[j];
		for (j = 0; j <= 10000; j++)
		{
			for (k = 0; k <= c[i] && j + k <= 10000; k++)
			{
				t[j + k] = (t[j + k] + old_t[j]) % MM;
			}
		}
		for (j = 0; j <= 10000; j++)
		{
			if (j >= pivot) old_t[j] = t[j];
			else old_t[j] = 0;
			t[j] = 0;
		}
	}

	ans = 0;
	for (i = 0; i <= 10000; i++)
	{
		ans = (ans + old_t[i]) % MM;
	}
	cout << ans << "\n";
}