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 <iostream>
#include <algorithm>

constexpr int N = 105;
constexpr int p = 1000000007;

int add(int x, int y) { return (x += y) >= p ? x - p : x; }
int sub(int x, int y) { return (x -= y) < 0 ? x + p : x; }

int n, q, k;
int c[N], b[N];
int f[N][N * N];
int s[2][N * N];
int l[N];

int main() {
    std::ios::sync_with_stdio(false), std::cin.tie(nullptr);
    std::cin >> n;
    int sumc = 0;
    for (int i = 1; i <= n; ++i) std::cin >> c[i], sumc += c[i];
    for (int i = 1; i < n; ++i) std::cin >> b[i];
    std::cin >> q;
    while (q--) {
        std::cin >> k;
        int sumb = k;
        for (int i = 1; i <= n; ++i) {
            l[i] = l[i - 1] + sumb;
            sumb += b[i];
        }
        f[0][0] = 1;
        for (int i = 0; i <= sumc; ++i) s[0][i] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < l[i] && j <= sumc; ++j) s[1][j] = 0;
            for (int j = std::max(l[i], 0); j <= sumc; ++j) {
                f[i][j] = add(f[i][j], sub(s[0][j], j - c[i] - 1 >= 0 ? s[0][j - c[i] - 1] : 0));
                s[1][j] = add(f[i][j], j > 0 ? s[1][j - 1] : 0);
            }
            std::swap(s[0], s[1]);
        }
        int min = k * n + sumb - k;
        int ans = 0;
        for (int i = std::max(min, 0); i <= sumc; ++i) {
            ans = add(ans, f[n][i]);
        }
        std::cout << ans << '\n';
    }
    return 0;
}
