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 INF = (int)1E9;
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int N;
int val[100005][19], a[100005];
int get(int l, int r) {
  int len = r - l + 1;
  for (int k = (0); k < (19); k++) {
    if ((1 << k) <= len && (1 << (k + 1)) > len) {
      len = k;
      break;
    }
  }
  int x = val[l][len];
  int y = val[r - (1 << len) + 1][len];
  return gcd(x, y);
}
int main() {
  cin >> N;
  map<int, long long> ans;
  for (int i = (0); i < (N); i++) {
    scanf("%d", a + i);
    val[i][0] = a[i];
  }
  for (int j = (1); j < (19); j++) {
    for (int i = (0); i < (N); i++) {
      if (i + (1 << j) > N) continue;
      val[i][j] = val[i][j - 1];
      int i2 = i + (1 << (j - 1));
      val[i][j] = gcd(val[i][j - 1], val[i2][j - 1]);
    }
  }
  for (int i = (0); i < (N); i++) {
    int g = a[i];
    int now = i;
    while (1) {
      int l = now + 1, r = N - 1;
      while (l <= r) {
        int m = (l + r) / 2;
        int gg = get(i, m);
        if (gg < g)
          r = m - 1;
        else
          l = m + 1;
      }
      ans[g] += l - now;
      now = l;
      if (now < N) {
        g = get(i, now);
      } else
        break;
    }
  }
  int Q;
  cin >> Q;
  while (Q--) {
    int x;
    scanf("%d", &x);
    printf("%I64d\n", ans[x]);
  }
  return 0;
}