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;
int n, q, v[100000], dp[100000][17], Log[100001];
map<int, long long> res;
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int GCD(int l, int r) {
  int len = Log[r - l + 1];
  return gcd(dp[l][len], dp[r - (1 << len) + 1][len]);
}
int main() {
  for (int i = 2; i <= 100000; ++i) Log[i] = Log[i / 2] + 1;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", v + i);
    dp[i][0] = v[i];
  }
  for (int k = 1; k < 17; ++k)
    for (int i = 0; i + (1 << k) <= n; ++i)
      dp[i][k] = gcd(dp[i][k - 1], dp[i + (1 << (k - 1))][k - 1]);
  for (int i = 0; i < n; ++i) {
    int j = i;
    int g = v[i];
    int prevLast = i - 1;
    while (true) {
      int l = j + 1, r = n - 1, last = j, m;
      while (l <= r) {
        m = (l + r) / 2;
        if (GCD(i, m) == g) {
          last = m;
          l = m + 1;
        } else
          r = m - 1;
      }
      res[g] += last - prevLast;
      j = last + 1;
      if (j >= n) break;
      prevLast = last;
      g = GCD(i, j);
    }
  }
  int x;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &x);
    printf("%lld\n", res[x]);
  }
  return 0;
}