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;
const double error = 1e-15;
const double pi = 2 * acos(0);
const int maxn = (int)1e5 + 10;
const int mod = (int)1e9;
long int gcd(long int a, long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long v[100005];
vector<pair<long, long long> > vp1, vp;
int main() {
  long n, flg, i, j, mm, q, g;
  map<long, long long> m;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> v[i];
  }
  m[v[0]]++;
  vp1.push_back(make_pair(v[0], 1));
  for (i = 1; i < n; i++) {
    if (i % 2 == 1) {
      for (j = 0; j < vp1.size(); j++) {
        g = gcd(vp1[j].first, v[i]);
        m[g] += vp1[j].second;
        if (vp.size() > 0 && vp[vp.size() - 1].first == g) {
          vp[vp.size() - 1].second += vp1[j].second;
        } else
          vp.push_back(make_pair(g, vp1[j].second));
      }
      m[v[i]]++;
      vp.push_back(make_pair(v[i], 1));
      vp1.clear();
      sort(vp.begin(), vp.end());
    } else {
      for (j = 0; j < vp.size(); j++) {
        g = gcd(vp[j].first, v[i]);
        m[g] += vp[j].second;
        if (vp1.size() > 0 && vp1[vp1.size() - 1].first == g) {
          vp1[vp1.size() - 1].second += vp[j].second;
        } else
          vp1.push_back(make_pair(g, vp[j].second));
      }
      m[v[i]]++;
      vp1.push_back(make_pair(v[i], 1));
      vp.clear();
      sort(vp1.begin(), vp1.end());
    }
  }
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> j;
    cout << m[j] << endl;
  }
  return 0;
}