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.
/*
F(a,b) = min((a[i]前缀-b[i-1]前缀)/i)
现今b全部确定，须使a[i]前缀必大于某一值。 
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,c[200],b[200],l[200],f[110][20000],q,x,pb1,pb2;
int main(){
	cin >> n;
	for(ll i=1;i<=n;i++)scanf("%lld",&c[i]);
	for(ll i=1;i<n;i++)scanf("%lld",&b[i]);
	cin >> q;
	while(q--){
		cin >> x;
		pb2=pb1=b[1];
		for(ll i=2;i<=n;i++){
			l[i]=x*i+pb2;
			pb1 += b[i];
			pb2 += pb1;
			//printf("l[%lld]=%lld\n",i,l[i]);
		}
		l[1]=x;
		f[0][0]=1;
		for(ll i=1;i<=n;i++){
			for(ll k=0;k<=10000;k++){
				for(ll j=0;j<=c[i];j++){
					if((k+j>=l[i]))f[i][k+j]=(f[i-1][k]+f[i][k+j])%1000000007;
					//if(k<=10)printf("f[%lld][%lld] = %lld\n",i,k+j,f[i][k+j]);
				}			
			}
		}
		ll ans=0;
		for(ll k=0;k<=10000;k++){
			ans += f[n][k];
			ans %= 1000000007;
		}
		cout << ans;
	}
}