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>
#define BPhuong "main"
#define maxn 109
#define fi(i, a, b) for(int i = a; i <= b; i++)
#define fid(i, a, b) for(int i = a; i >= b; i--)
#define ll long long
#define bit(i, j) ((i >> j) & 1)
#define pii pair<int, int>
#define f first
#define se second
#define pb push_back

using namespace std;
int n, q, c[maxn], b[maxn];
int dp[maxn][maxn * maxn], ans[200008], F[maxn * maxn];
const int mod = 1e9 + 7;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
    #ifndef ONLINE_JUDGE
    freopen(BPhuong ".inp", "r", stdin);
    freopen(BPhuong ".out", "w", stdout);
    #endif // ONLINE_JUDGE
    cin >> n;
    ll gt = 1;
    fi(i, 1, n){
        int u;
        cin >> u;
        c[i] = c[i - 1] + u;
        gt = gt * (u + 1) % mod;
    }
    c[n + 1] = c[n];
    fi(i, 1, n - 1){
        int u;
        cin >> u;
        b[i] = b[i - 1] + u;
    }
    fi(i, 2, n - 1) b[i] += b[i - 1];
    fi(t, -100000, 100000){
        if(t * n + b[n - 1] <= 0) {
                ans[t + 100000] = gt;
                continue;
        }
        if(n * t + b[n - 1] > c[n]) {
                ans[t + 100000] = 0;
                continue;
        }
        fi(i, 1, n)
            fi(j, 0, 10000) dp[i][j] = 0;
        fi(i, max(0, t), c[1]) dp[1][i] = 1;
        fi(j, 1, c[1 + 1])
                dp[1][j] = (dp[1][j] + dp[1][j - 1]) % mod;
        fi(i, 2, n){
            fi(j, max(t * i + b[i - 1], 0), c[i])
                if(j - c[i] + c[i - 1] - 1 >= 0)
                    dp[i][j] = (dp[i - 1][j] - dp[i - 1][j - c[i] + c[i - 1] - 1] + mod) % mod;
                else dp[i][j] = dp[i - 1][j];
            fi(j, 1, c[i + 1])
                dp[i][j] = (dp[i][j] + dp[i][j - 1]) % mod;
        }
        ans[t + 100000] = dp[n][c[n]];
    }
    cin >> q;
    while(q--){
        int x;
        cin >> x;
        cout << ans[x + 100000] << '\n';
    }

}
