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 sz(s) (int)s.size()
#define f first
#define s second
#define pb push_back
#define all(s) s.begin(), s.end()

#define vi vector<int>
#define vvi vector<vi>

#define ll long long
#define vll vector<ll>

#define ii pair<int, int>
#define vii vector<ii>
#define vvii vector<vii>

const int INF = 1e9 + 7;
const int mod = INF;
const int N = 1e6 + 7;
const int T = 1 << 20;

int c[N];
int b[N];
int d[N];

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

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

    int q;
    cin >> q;
    int shift = 100 * 100 + 7;

    for(int _ = 0; _ < q; _++){
        int zap;
        cin >> zap;
        // policz schodki
        d[0] = zap;
        int last = d[0];
        for(int i= 0; i < n - 1; i++){
            last += b[i];
            d[i + 1] = d[i] + last;
            d[i + 1] = min(d[i + 1], shift - 1);
        }

        vi dp(shift + shift);
        dp[shift] = 1;

        for(int i = 0; i < n; i++){
            vi dp2(shift + shift);
            for(int j = 0; j <= c[i]; j++)
                for(int k = 0; k < shift + shift; k++)
                    if(k + j < shift + shift and k + j >= d[i] + shift)
                        dp2[k + j] = (dp2[k + j] + dp[k]) % INF; 
            swap(dp, dp2);
        }
        
        int ans = 0;
        for(int i = 0; i < shift + shift; i++)
            ans = (ans + dp[i]) % INF; 
        cout << ans << '\n';
    }
}

signed main(){
    int test = 1;
    // cin >> test;

    for(int i = 0; i < test; i++)
        solve();
}