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 sz(a) (int)(a).size()
#define rep(i, a, b) for (int i = (a), _b = (b); i < _b; ++i)
#define debug(a) cerr << #a << ": " << (a) << '\n';
#define debug_arr(a, n) { cerr << #a << " "; rep(i, 0, n) cerr << a[i] << " \n"[(i + 1) == n]; }
#define debug_two(a, b) cerr << #a << ": " << (a) << ", " << #b << ": " << (b) << '\n';

template <typename num_t>
inline void add_mod(num_t& a, const long long& b, const int& m) { a = (a + b) % m; if (a < 0) a += m; }
template <typename num_t>
inline bool update_max(num_t& a, const num_t& b) { return a < b ? a = b, true : false; }
template <typename num_t>
inline bool update_min(num_t& a, const num_t& b) { return a > b ? a = b, true : false; }
template <typename num_t>
num_t gcd(num_t lhs, num_t rhs) { return !lhs ? rhs : gcd(rhs % lhs, lhs); }
template <typename num_t>
num_t pw(num_t n, num_t k, num_t mod) {
  num_t res = 1; for (; k > 0; k >>= 1) { if (k & 1) res = 1ll * res * n % mod; n = 1ll * n * n % mod; } return res;
}
int invserse(int n, int mod) {
  return pw(n, mod - 2, mod);
}

typedef long long int64;
const int mod = 1e9 + 7;

void solve() {
  int n;
  cin >> n;
  vector<int> c(n); rep(i, 0, n) cin >> c[i];
  vector<int> b(n, 0);
  rep(i, 1, n) { 
    cin >> b[i];
    b[i] += b[i - 1];
  }
  vector<int> sum_b(n, 0);
  rep(i, 0, n) {
    sum_b[i] = b[i];
    if (i) sum_b[i] += sum_b[i - 1];
  }

  const int base = 10007;
  map<int, int> pre_calc;

  int sum_c = accumulate(c.begin(), c.end(), 0);

  // debug_two(n, mx_a - sum_b.back());
  // debug(-sum_b.back() / n);

  function<int(int)> get_value = [&](int x) -> int {
    if (x > c[0])
      return 0;

    if (pre_calc.count(x)) 
      return pre_calc[x];

    if (x < 0 && sum_b.back() == 0) {
      return get_value(0);
    }

    if (x < -sum_b.back() / n - 1) {
      return get_value(-sum_b.back() / n - 1);
    }
  
    if (x < 0 && (sum_c - sum_b.back()) / n < x) {
      // debug_two(x, (sum_c - sum_b.back()) / n)
      return get_value((sum_c - sum_b.back()) / n);
    }

    vector<int> pre, cur;
    cur.assign(base, 0);
    rep(val, max(0, x), c[0] + 1) {
      cur[val] = 1;
    }
    rep(i, 1, sz(cur)) cur[i] += cur[i - 1];

    rep(i, 1, n) {
      pre = cur;
      fill(cur.begin() + max(0, i * x + sum_b[i - 1]), cur.end(), 0);

      rep(sum, max(0, (i + 1) * x + sum_b[i]), sz(cur)) {
        cur[sum] = pre[sum];
        if (sum - c[i] - 1 >= 0) {
          add_mod(cur[sum], -pre[sum - c[i] - 1], mod);
        }
        if (sum)
          add_mod(cur[sum], cur[sum - 1], mod);
      }
    }

    pre_calc[x] = cur.back();
    return cur.back();
  };

  int q = 0; cin >> q;
  while (q-- > 0) {
    int x; cin >> x;
    cout << get_value(x) << '\n';
  }
}

int main() {
  ios_base::sync_with_stdio(0); cin.tie(0);
#ifdef _LOCAL_
  freopen("in.txt", "r", stdin);
#endif
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test)
    cin >> test;

  for(int i = 0; i < test; ++i) {
    //printf("Case #%d: ", i + 1);
    solve();
  }
#ifdef _LOCAL_
  cerr << "\n" << 1.0 * clock() / CLOCKS_PER_SEC << "\n";
#endif
}