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 int N = 1e5;
int n, q, a[N + 5];
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int f[N + 5][20];
struct Node {
  long long i, cnt;
  bool operator<(const Node& x) const { return i < x.i; }
};
vector<Node> v, vv;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    f[i][0] = a[i];
  }
  for (int k = 1; (1ll << k) <= n; k++)
    for (int i = 1; i + (1ll << k) - 1 <= n; i++)
      f[i][k] = gcd(f[i][k - 1], f[i + (1ll << (k - 1))][k - 1]);
  for (int i = 1; i <= n; i++) {
    int L = i, GCD = a[i];
    while (true) {
      int ans = 0, l = 1, r = L;
      while (l <= r) {
        int mid = (l + r) >> 1;
        int k = log2(i - mid + 1);
        int s = gcd(f[mid][k], f[i - (1ll << k) + 1][k]);
        if (s == GCD)
          ans = mid, r = mid - 1;
        else
          l = mid + 1;
      }
      Node New;
      New.i = GCD, New.cnt = L - ans + 1;
      v.push_back(New);
      L = ans - 1;
      if (L < 1) break;
      GCD = gcd(GCD, a[L]);
    }
  }
  sort(v.begin(), v.end());
  Node New;
  New.i = v[0].i, New.cnt = v[0].cnt;
  vv.push_back(New);
  for (int i = 1; i < v.size(); i++) {
    if (v[i].i == v[i - 1].i)
      vv[vv.size() - 1].cnt += v[i].cnt;
    else {
      Node Ne;
      Ne.i = v[i].i, Ne.cnt = v[i].cnt;
      vv.push_back(Ne);
    }
  }
  scanf("%d", &q);
  while (q--) {
    int x;
    scanf("%d", &x);
    long long l = 0, r = vv.size() - 1, ans = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (vv[mid].i == x) {
        ans = vv[mid].cnt;
        break;
      }
      if (vv[mid].i > x)
        r = mid - 1;
      else
        l = mid + 1;
    }
    printf("%lld\n", ans);
  }
  return 0;
}