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:
/*
K.D. Vinit  |,,|
*/
#include<bits/stdc++.h>
using namespace std;

#define int long long
const int mod = 1e9 + 7;
const int W = 1e4 + 1;

void solve()
{
    int n;
    cin>>n;

    int c[n+1], b[n+1];

    c[0]=0;
    for(int i=1; i<=n; i++) cin>>c[i];

    b[1]=0;
    for(int i=2; i<=n; i++) cin>>b[i];
    for(int i=2; i<=n; i++) b[i]+=b[i-1];       //Prefix Sum of b[i];
    for(int i=2; i<=n; i++) b[i]+=b[i-1];       //Prefix sum of Pb[i];

    int q, x;
    cin>>q>>x;

    int dp[W][n+1];
    memset(dp, 0, sizeof(dp));

    dp[0][0]=1;

    for(int i=1; i<=n; i++)
    {
        int l = (i*x + b[i]);
        for(int j=W-1; j>=0; j--)
        {
            if(dp[j][i-1]==0) continue;
            int low = max((int)0, l-j);
            for(int w=c[i]; w>=low; w--) dp[j+w][i]+=dp[j][i-1], dp[j+w][i]%=mod;
        }
    }

    int ans=0;
    for(int i=0; i<W; i++) ans+=dp[i][n], ans%=mod;
    cout<<ans<<endl;
}

int32_t main()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    solve();
    return 0;
}