Problem: Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.

<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.

Input

The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).

The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).

Output

For each query print the result in a separate line.

Examples

Input

3
2 6 3
5
1
2
3
4
6


Output

1
2
2
0
1


Input

7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000


Output

14
0
2
2
2
0
2
2
1
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
constexpr long long MOD = 1e9 + 7;
int jmp[100100][25];
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int G(int i, int j) {
  int k = 0;
  while ((1 << (k + 1)) <= j - i + 1) ++k;
  int ans = gcd(jmp[i][k], jmp[j - (1 << k) + 1][k]);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> jmp[i][0];
  for (int k = 1; k <= 20; ++k) {
    for (int i = 0; i < n; ++i) {
      int j = i + (1 << (k - 1));
      if (j >= n)
        jmp[i][k] = jmp[i][k - 1];
      else
        jmp[i][k] = gcd(jmp[i][k - 1], jmp[j][k - 1]);
    }
  }
  map<int, long long> m;
  for (int i = 0; i < n; ++i) {
    int j = i;
    int cur = jmp[i][0];
    while (j < n) {
      int lo = j, hi = n - 1, mid, upper;
      int val = G(i, j);
      while (lo <= hi) {
        mid = (lo + hi) / 2;
        if (G(i, mid) >= val) {
          lo = mid + 1;
          upper = mid;
        } else {
          hi = mid - 1;
        }
      }
      m[val] += upper - j + 1;
      j = upper + 1;
    }
  }
  int nq;
  cin >> nq;
  while (nq--) {
    int x;
    cin >> x;
    cout << m[x] << endl;
  }
  return 0;
}