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;

const int N = 110;
const int MOD = (int)1e9 + 7;

inline int add(int u, int v, int mod = MOD) {
  return (u += v) >= mod ? u - mod : u;
}
inline int sub(int u, int v, int mod = MOD) {
  return (u -= v) < 0 ? u + mod : u;
}
inline int mul(int u, int v, int mod = MOD) { return (long long)u * v % mod; }
inline int power(int u, int v, int mod = MOD) {
  int res = 1;
  for (; v; v >>= 1, u = mul(u, u, mod))
    if (v & 1) res = mul(res, u, mod);
  return res;
}
inline int inv(int u, int mod = MOD) { return power(u, mod - 2, mod); }
inline void add_to(int &u, int v, int mod = MOD) { u = add(u, v, mod); }
inline void sub_to(int &u, int v, int mod = MOD) { u = sub(u, v, mod); }
inline void mul_to(int &u, int v, int mod = MOD) { u = mul(u, v, mod); }

int n;
int a[N];
int b[N];
int sum_b[N];
int sum_ib[N];

int q;
int x[N];

int f[N][N * N];

int main() {
  ios_base::sync_with_stdio(0); cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    cin >> b[i];
  }
  for (int i = 1; i < n; i++) {
    sum_b[i] = sum_b[i - 1] + b[i];
    sum_ib[i] = sum_ib[i - 1] + i * b[i];
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> x[i];
  }
  f[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int prev_sum = 0; prev_sum < N * N; prev_sum++) {
      if (f[i - 1][prev_sum]) {
        for (int j = 0; j <= a[i]; j++) {
          int sum = prev_sum + j;
          int now = sum - sum_b[i - 1] * i + sum_ib[i - 1];
          if (now >= x[1] * i) {
            add_to(f[i][sum], f[i - 1][prev_sum]);
          }
        }
      }
    }
  }
  int res = 0;
  for (int i = 0; i < N * N; i++) {
    add_to(res, f[n][i]);
  }
  cout << res << '\n';
  return 0;
}