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 int long long
using namespace std;

// https://github.com/nweeks1/UltiSnips
template <const int32_t MOD> struct ModInt {
  int32_t x;
  ModInt() : x(0) {}
  ModInt(long long u) : x(u) {
    x %= MOD;
    if (x < 0)
      x += MOD;
  }
  friend bool operator==(const ModInt &a, const ModInt &b) {
    return a.x == b.x;
  }
  friend bool operator!=(const ModInt &a, const ModInt &b) {
    return a.x != b.x;
  }
  friend bool operator<(const ModInt &a, const ModInt &b) { return a.x < b.x; }
  friend bool operator>(const ModInt &a, const ModInt &b) { return a.x > b.x; }
  friend bool operator<=(const ModInt &a, const ModInt &b) {
    return a.x <= b.x;
  }
  friend bool operator>=(const ModInt &a, const ModInt &b) {
    return a.x >= b.x;
  }
  static ModInt sign(long long k) {
    return ((k & 1) ? ModInt(MOD - 1) : ModInt(1));
  }

  ModInt &operator+=(const ModInt &m) {
    x += m.x;
    if (x >= MOD)
      x -= MOD;
    return *this;
  }
  ModInt &operator-=(const ModInt &m) {
    x -= m.x;
    if (x < 0LL)
      x += MOD;
    return *this;
  }
  ModInt &operator*=(const ModInt &m) {
    x = (1LL * x * m.x) % MOD;
    return *this;
  }

  friend ModInt operator-(const ModInt &a) {
    ModInt res(a);
    if (res.x)
      res.x = MOD - res.x;
    return res;
  }
  friend ModInt operator+(const ModInt &a, const ModInt &b) {
    return ModInt(a) += ModInt(b);
  }
  friend ModInt operator-(const ModInt &a, const ModInt &b) {
    return ModInt(a) -= ModInt(b);
  }
  friend ModInt operator*(const ModInt &a, const ModInt &b) {
    return ModInt(a) *= ModInt(b);
  }

  static long long fp(long long u, long long k) {
    long long res = 1LL;
    while (k > 0LL) {
      if (k & 1LL)
        res = (res * u) % MOD;
      u = (u * u) % MOD;
      k /= 2LL;
    }
    return res;
  }

  static constexpr int mod() { return MOD; }

  ModInt fastpow(long long k) { return ModInt(fp(x, k)); }
  ModInt inv() {
    assert(x);
    return ModInt(fp(x, MOD - 2));
  }
  ModInt &operator/=(const ModInt &m) { return *this *= ModInt(m).inv(); }
  friend ModInt operator/(const ModInt &a, const ModInt &b) {
    return ModInt(a) *= ModInt(b).inv();
  }

  friend ostream &operator<<(ostream &out, const ModInt &a) {
    return out << a.x;
  }
  friend istream &operator>>(istream &in, ModInt &a) { return in >> a.x; }
};

const int MOD = 1e9 + 7;
using Mint = ModInt<MOD>;

const int MAXSUM = 1e4 + 1;
const int MAXN = 101;

Mint ways[MAXN][MAXSUM];

signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;

  vector<int> c(n);
  vector<int> b(n - 1);
  for (int i = 0; i < n; ++i)
    cin >> c[i];
  for (int i = 0; i < n - 1; ++i)
    cin >> b[i];
  vector<int> partialB(n);
  for (int k0 = 0; k0 < n; ++k0)
    for (int i = 0; i < k0; ++i)
      partialB[k0] += (k0 - i) * b[i];

  int q;
  cin >> q;
  int x;
  cin >> x;

  ways[0][0] = 1;
  for (int nbPris = 1; nbPris <= n; ++nbPris)
    for (int curSum = 0; curSum < MAXSUM; ++curSum) {
      if (curSum < nbPris * x + partialB[nbPris - 1])
        continue;
      for (int v = 0; v <= c[nbPris - 1] and v <= curSum; ++v)
        ways[nbPris][curSum] += ways[nbPris - 1][curSum - v];
    }

  Mint sol = 0;
  for (int sum = 0; sum < MAXSUM; ++sum)
    sol += ways[n][sum];
  cout << sol << endl;
}