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;
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tiii;
typedef tuple<ll, ll, ll> tlll;
const int INF = 100000000;
const ll INFINITE = 4500000000000000000;
typedef pair<int, char> pic;
int dx[4] = { -1, 1, 0, 0 };
int dy[4] = { 0, 0, 1, -1 };

const int MOD = 1e9 + 7;

int main(void) {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

	int n; cin >> n;
	vector<int> c(n + 1);
	vector<int> pre(n + 1);
	for (int i = 1; i <= n; i++) {
		cin >> c[i];
	}
	vector<int> b(n);
	for (int i = 1; i <= n - 1; i++) {
		cin >> b[i];
		pre[i] = pre[i - 1] + b[i];
	}
	vector<ll> answers(201);
	//for (int x = -100; x <= 100; x++) {
	//	vector<int> limit(n + 1);
	//	limit[1] = x;
	//	for (int i = 2; i <= n; i++) {
	//		limit[i] = limit[i - 1] + x + pre[i - 1];
	//	}
	//	vector<vector<ll>> world(n + 1, vector<ll>(10004));
	//	world[0][0] = 1;
	//	for (int i = 1; i <= n; i++) {
	//		for (int j = 0; j < 10004; j++) {
	//			if (world[i - 1][j] == 0) continue;
	//			for (int k = 0; k <= c[i]; k++) {
	//				if (j + k >= limit[i]) {
	//					world[i][j + k] += world[i - 1][j];
	//					world[i][j + k] %= MOD;
	//				}
	//			}
	//		}
	//	}
	//	ll ans = 0;
	//	for (int i = max(0, limit[n]); i < 10004; i++) {
	//		ans += world[n][i];
	//		ans %= MOD;
	//	}
	//	answers[x + 100] = ans;
	//}
	int q; cin >> q;
	for (int i = 1; i <= q; i++) {
		int query;
		cin >> query;
		int x = query;
		vector<int> limit(n + 1);
		limit[1] = x;
		for (int i = 2; i <= n; i++) {
			limit[i] = limit[i - 1] + x + pre[i - 1];
		}
		vector<vector<ll>> world(n + 1, vector<ll>(10004));
		world[0][0] = 1;
		for (int i = 1; i <= n; i++) {
			for (int j = 0; j < 10004; j++) {
				if (world[i - 1][j] == 0) continue;
				for (int k = 0; k <= c[i]; k++) {
					if (j + k >= limit[i]) {
						world[i][j + k] += world[i - 1][j];
						world[i][j + k] %= MOD;
					}
				}
			}
		}
		ll ans = 0;
		for (int i = max(0, limit[n]); i < 10004; i++) {
			ans += world[n][i];
			ans %= MOD;
		}
		cout << ans << endl;
	}

}