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>

using namespace std;

#define forn(I,N) for (int I = 0; I < N; ++I)
#define fornd(I,N) for (int I = N-1; I >= 0; --I)
#define forab(I,A,B) for (int I = A; I <= B; ++I)
#define forabd(I,A,B) for (int I = B; I >= A; --I)
#define foreach(I, A) for(typeof((A).begin()) I = (A).begin(); I != (A).end(); ++I)
// #define foreach(I, A) for((__typeof__(A))::iterator I = (A).begin(); I != (A).end(); ++I)

#define pb push_back
#define mp make_pair

typedef long long LL;

// sorting
struct P {
  int x, y;
  bool operator<(const P &p) {
    if (x != p.x) return x < p.x;
    else return y < p.y;
  }
};

/*
bool comp(data dt1, data dt2) {
  return dt1 < dt2;
}
*/

const LL mod = 1e9 + 7;

int main() {
  int n;
  cin >> n;

  vector<int> c(n), b(n);
  forn (i, n) {
    cin >> c[i];
  }
  forn (i, n - 1) {
    cin >> b[i + 1];
  }

  int q;
  cin >> q;

  forn (t, q) {
    LL x;
    cin >> x;

    LL s1 = 0, s2 = 0;
    vector<LL> ct;
    ct.pb(1);
    forn (i, n) {
      s1 = (s1 + b[i]) % mod;
      s2 = (s2 + s1) % mod;
      vector<LL> tct(ct.size() + c[i]);

      /*
      forab (j, 0, c[i]) {
        forn (k, ct.size()) {
          if ((LL)j + (LL)k - s2 >= (LL)(i + 1) * x) {
            tct[j + k] = (tct[j + k] + ct[k]) % mod;
          }
        }
      }
      */

      LL st = 0;
      forn (j, ct.size() + c[i]) {
        if (j < ct.size()) {
          st = (st + ct[j]) % mod;
        }
        if (j - c[i] - 1 >= 0) {
          st = (st + mod - ct[j - c[i] - 1]) % mod;
        }
        if ((LL)j - s2 >= (LL)(i + 1) * x) {
          tct[j] = st;
        }
      }

      ct = tct;
    }

    LL ans = 0;
    forn (i, ct.size()) {
      ans = (ans + ct[i]) % mod;
    }
    cout << ans << endl;
  }

  return 0;
}