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>
#define pii pair<int,int>
#define fi first
#define sc second
#define pb push_back
#define ll long long
#define trav(v,x) for(auto v:x)
#define all(x) (x).begin(), (x).end()
#define VI vector<int>
#define VLL vector<ll>
using namespace std;
const int N = 1e6 + 100;
const ll mod = 1e9 + 7;

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
    int n;
    cin >> n;
    VI mx(n + 1);
    for(int i = 1; i <= n; i++)
        cin >> mx[i];
    VI b(n + 1);
    for(int i = 1; i < n; i++)
        cin >> b[i], b[i] += b[i - 1];
    for(int i = 1; i < n; i++)
        b[i] += b[i - 1];
    vector<VLL> dp(n + 1, VLL(10010, 0));
//    VLL ans(300);
    ll ans = 0;
    int x;
    cin >> x;
    cin >> x;
  //  for(int x = -101; x <= 101; x++)
    {
        dp[0][0] = 1;
        int suma = 0;
        for(int i = 1; i <= n; i++)
        {
            fill(all(dp[i]), 0);
            for(int val = 0; val <= mx[i]; val++)
            {
                int mn = b[i - 1] + x * i - val;
                mn = max(mn, 0);
                for(int bf = mn; bf <= suma; bf++)
                {
                    dp[i][val + bf] += dp[i - 1][bf];
                    if(dp[i][val + bf] >= mod)dp[i][val + bf] -= mod;
                }
            }
            suma += mx[i];
        }
//        ans[x + 110] = 0;
//        for(int i = 0; i <= suma; i++)
//        {
//            ans[x + 110] += dp[n][i];
//            if(ans[x + 110] >= mod)ans[x + 110] -= mod;
//        }
            ans = 0;
            for(int i = 0; i <= suma; i++)
        {
            ans+= dp[n][i];
            if(ans >= mod)ans -= mod;
        }
        cout << ans << '\n';
    }
//    int tt;
//    cin >> tt;
//    while(tt --)
//    {
//        int x;
//        cin >> x;
//        x = max(x, -101);
//        x = min(x, 101);
//        cout << ans[x + 110] << '\n';
//    }
}