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>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ll MOD = 1e9+7;

int N, Q;
ll C[110], CS[110];
ll B[110];
ll T[110];
ll X[100100];
ll D[110][10100];

ll solve(ll X) {
	memset(D,0,sizeof(D));
	D[0][0] = 1;
	ll base = 0;
	for (int i=0;i<N;i++) {
		for (ll j=max(-base, 0LL);j<=CS[i];j++) {
			for (int k=0;k<=C[i+1];k++) {
				D[i+1][j+k] = (D[i+1][j+k]+D[i][j])%MOD;
			}
		}
		base -= T[i]+X;
	}
	ll ans = 0;
	for (ll j=max(-base,0LL);j<=CS[N];j++) ans = (ans+D[N][j])%MOD;
	return ans;
}

int main () {
	scanf("%d",&N);
	for (int i=1;i<=N;i++) scanf("%lld",&C[i]);
	for (int i=1;i<=N;i++) CS[i] = CS[i-1]+C[i];
	for (int i=1;i<=N-1;i++) scanf("%lld",&B[i]);
	for (int i=1;i<=N-1;i++) T[i] = T[i-1]+B[i];
	scanf("%d",&Q);
	for (int i=0;i<Q;i++) scanf("%lld",&X[i]);
	for (int i=0;i<Q;i++) {
		printf("%lld\n",solve(X[i]));
	}

	return 0;
}