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>
using namespace std;
#define int long long
const int mod=1e9+7;
int mul(int x,int y)
{
    return x*y%mod;
}
int add(int x,int y)
{
    x+=y;
    if(x>=mod)
    x-=mod;
    return x;
}
int sub(int x,int y)
{
    x-=y;
    if(x<0)
    x+=mod;
    return x;
}
void solve()
{
    int n;
    cin>>n;
    vector<int> a(n),b(n-1),h(n);
    for(auto& x:a)
    cin>>x;
    for(auto& x:b)
    cin>>x;
    {
        int res=0,cur=0;
        for(int i=0;i<n-1;i++)
        {
            cur+=b[i];
            cur%=mod;
            res+=cur;
            res%=mod;
            h[i+1]=res;
        }
    }
    int q;
    cin>>q;
    while(q--)
    {
        int ty;
        cin>>ty;
        int tot=accumulate(a.begin(),a.end(),0LL)+10;
        tot%=mod;
        vector<int> dp(tot);
        dp[0]=1;
        for(int k=0;k<n;k++)
        {
            int v=a[k];
            for(int i=tot-v-2;i>=0;i--)
            dp[i+v+1]=sub(dp[i+v+1],dp[i]);
            for(int i=0;i<tot-1;i++)
            dp[i+1]=add(dp[i+1],dp[i]);
            int z=mul(ty,k+1)+h[k];
            z%=mod;
            z=min(z,tot);
            for(int j=0;j<z;j++)
            dp[j]=0;
        }
        cout<<accumulate(dp.begin(),dp.end(),0LL)%mod<<'\n';
    }
}
int32_t main()
{
    int t=1;
    //cin>>t;
    while(t--)
    solve();
    return 0;
}
