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:
/**
  * @brief codeforces
  * @author yao
  */
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <utility>
#include <algorithm>
#include <functional>
#include <climits>
#define ft first
#define sd second
#ifdef DBG
#   define dbg_pri(x...) fprintf(stderr,x)
#else
#   define dbg_pri(x...) 0
#   define NDEBUG
#endif //DBG
#include <cassert>

typedef unsigned int uint;
typedef long long int lli;
typedef unsigned long long int ulli;

#define N 128
#define P ((int)1e9+7)

int c[N];
int b[N];
int s[N];
int dp[N*N];

int main()
{
    int n;
    scanf("%d", &n);
    for(int i=0;i<n;++i) scanf("%d", &c[i]);
    for(int i=0;i<n-1;++i) scanf("%d", &b[i]);
    int q;
    scanf("%d", &q);
    for(int qc=0; qc<q; ++qc)
    {
        int x;
        scanf("%d", &x);
        s[0] = x;
        for(int i=1;i<n;++i) s[i] = s[i-1]+b[i-1];
        memset(dp,0,sizeof(dp));
        dp[0] = 1;
        int tar = 0;
        for(int i=0;i<n;++i)
        {
            for(int j=i*100,ej=std::max(0,tar);j>=ej;--j)if(dp[j])
            {
                for(int k=j+1,ek=j+c[i];k<=ek;++k)
                    dp[k] = (dp[k] + dp[j]) % P;
            }
            tar += s[i];
        }
        int ans = 0;
        for(int i=std::max(0,tar),e=n*100;i<=e;++i)
            ans = (ans + dp[i]) %P;
        printf("%d\n", ans);
    }
    return 0;
}