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:
#include <bits/stdc++.h>
using namespace std;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef long long ll;
typedef long double ld;
#define INF 2001001001
#define MOD 1000000007

ll N,X,C[105],B[105],D[105],D1[105];
int Q;
ll dp[105][10005];
ll C2[105];

int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);
  cin>>N;
  for (int i=1;i<=N;i++)
    cin>>C[i];
  for (int i=1;i<N;i++)
    cin>>B[i];
  cin>>Q;
  cin>>X;
  D1[1]=X;
  for (int i=2;i<=N;i++)
    D1[i]=D1[i-1]+B[i-1];
  for (int i=1;i<=N;i++){
    D[i]=D[i-1]+D1[i];
    C2[i]=C2[i-1]+C[i];
  }
  dp[0][0]=1;
  for (int i=0;i<N;i++){ //adding in element i+1
    for (int j=0;j<=C2[i];j++){
      if (dp[i][j]==0) continue;
      for (int k=0;k<=C[i+1];k++){
	if (j+k>=D[i+1]){
	  dp[i+1][j+k]+=dp[i][j];
	  if (dp[i+1][j+k]>=MOD)
	    dp[i+1][j+k]%=MOD;
	}
      }
    }
  }
  ll ans=0;
  for (int i=0;i<=10000;i++)
    ans=(ans+dp[N][i])%MOD;
  cout<<ans<<endl;
  return 0;
}