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>
#define ll long long
using namespace std;
const ll maxn=205;
const ll cn0=0;
const ll mod=1e9+7;
ll b[maxn],c[maxn],bp[maxn],bpp[maxn],cp[maxn],dp[maxn][maxn*maxn];
int n;
int main(){
	cin>>n;bp[0]=bpp[0]=cp[0]=0; 
	for(int i=1;i<=n;i++){scanf("%lld",&c[i]);cp[i]=c[i]+cp[i-1];}
	for(int i=1;i<n;i++){scanf("%lld",&b[i]);bp[i]=bp[i-1]+b[i];bpp[i]=bpp[i-1]+bp[i];}
	memset(dp,0,sizeof(dp));int q;cin>>q;int x;cin>>x;
	for(int i=max(0,x);i<=c[1];i++)dp[1][i]=1;
	for(ll i=2;i<=n;i++){
		for(ll j=max(cn0,x*i+bpp[i-1]);j<=cp[i];j++){
			for(ll k=max(cn0,j-c[i]);k<=j;k++){
				dp[i][j]=(dp[i][j]+dp[i-1][k])%mod;
			}
		}
	}
	ll ans=0;
	for(int i=0;i<=cp[n];i++)ans=(ans+dp[n][i])%mod;
	printf("%lld\n",ans);
}
