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.
//MY code
#include <bits/stdc++.h>
#define ll long long int
#define mp make_pair
#define pb push_back
#define ff first
#define ss second
#define mod 1000000007
using namespace std;
ll dp[101][20004], c[101], b[101],btot[101], q, x, n, bsum;
ll func(int pos, int sub)
{
    //cout<<pos<<" "<<sub<<"\n";
    if(pos==n)
    {
        return 1;
    }
    
    if(dp[pos][sub]!=-1)
    {
        return dp[pos][sub];
    }
    ll res=0;
    for(int i=0;i<=c[pos];i++)
    {
        //if(i-b[pos-1]>=0)
        if(sub+i>=x*(pos+1)+btot[pos])
        res = (res%mod + func(pos+1, sub+i)%mod)%mod;        
    }
    return dp[pos][sub]=res%mod;
}
 
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	ll ans=1;
	bsum=0;
	//float c[101], b[101];
	cin>>n;
	for(int i=0;i<n;i++)
	{
	    cin>>c[i];
	    ans = ((ans%mod) * (c[i]+1)%mod)%mod;
	}
	btot[0]=0;
	for(int i=1;i<n;i++)
	{
	    cin>>b[i];
	    bsum+=b[i];
	    btot[i]=btot[i-1]+bsum;
	}
	//cout<<bsum<<"\n";
	cin>>q;
	for(int i=0;i<q;i++)
	{
	    cin>>x;
	    for(int i=0;i<101;i++)
    	{
    	    for(int j=0;j<20004;j++)
    	    {
    	        //for(int k=0;k<101;k++)
    	        dp[i][j]=-1;
    	    }
    	}
	    ll res=func(0,0)%mod;
    	cout<<res<<"\n";
    	//cout<<(ans%mod-res%mod + mod)%mod<<"\n";
	}
	return 0;
}