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;
template <typename T>
vector<int> readVector(int n) {
  vector<T> res(n);
  for (int i = 0; i < n; i++) cin >> res[i];
  return res;
}
int gcd(int a, int b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  auto v = readVector<int>(n);
  map<int, int> divisors;
  map<int, int> nextDivisors;
  map<int, long long int> results;
  for (int i = 0; i < n; i++) {
    nextDivisors.clear();
    for (auto &p : divisors) {
      nextDivisors[gcd(v[i], p.first)] += p.second;
    }
    nextDivisors[v[i]]++;
    swap(nextDivisors, divisors);
    for (auto &p : divisors) {
      results[p.first] += p.second;
    }
  }
  int q, x;
  cin >> q;
  while (q--) {
    cin >> x;
    cout << results[x] << endl;
  }
  return 0;
}