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 mod (int)(1e9+7)
using namespace std;

int n, t, q, x, i, j;
long long sum = 1;
int b[110], c[110], sp[110], f[10204], mp[1200000], sc[110], s[10204];

int main()
{
	cin >> n;
	for (i = 1; i <= n; i ++ ) cin >> c[i], sum = 1ll * sum * (c[i] + 1) % mod, sc[i] = sc[i - 1] + c[i];
	for (int i = 1; i < n; i ++ ) cin >> b[i], sp[i + 1] = sp[i] + b[i];
	for (int i = 2; i <= n; i ++ ) sp[i] += sp[i - 1];
	cin >> q;
	memset(mp, -1, sizeof mp);
	while (q -- )
	{
		cin >> x;
		if (mp[x + 500000] != -1) { cout << mp[x + 500000] << endl; continue; }
		for (i = 1; i <= n; i ++ ) if (x * i + sp[i] > 0) break;
		if (i > n) { cout << sum << endl; continue; }
		for (i = 1; i <= n; i ++ ) if (x * i + sp[i] > sc[i]) break;
		if (i <= n) { puts("0"); continue; }
		int l = 0; memset(f, 0, sizeof f);
		s[0] = f[0] = 1;
		for (i = 1; i <= n; i ++ )
		{
			int L = l;
			l = max(l, x * i + sp[i]);
			for (j = sc[i - 1] + 1; j <= sc[i]; j ++ ) s[j] = s[j - 1];
			for (j = sc[i]; j >= l; j -- ) f[j] = (s[j] - (j - c[i] - 1 >= L ? s[j - c[i] - 1] : 0)) % mod;
			if (l) s[l - 1] = 0;
			for (j = l; j <= sc[i]; j ++ ) s[j] = ((j ? s[j - 1] : 0) + f[j]) % mod;
		}
		mp[x + 500000] = (s[sc[n]] % mod + mod) % mod;
		cout << mp[x + 500000] << endl;
	}
	return 0;
}