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;
#pragma comment(linker, "/STACK:1024000000,1024000000")
void work();
int main() {
  work();
  return 0;
}
void scanf(int& x, char ch = 0) {
  while ((ch = getchar()) < '0' || ch > '9')
    ;
  x = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0';
}
int gcd(int a, int b) {
  if (a && b)
    while (a > b ? a = a % b : b = b % a)
      ;
  return a + b;
}
int a[111111];
int q[333333];
int g[333333];
long long ans[333333];
void work() {
  int n, m;
  while (scanf("%d", &n) == 1) {
    for (int i = (1); i <= (n); i++) scanf(a[i]);
    int m;
    scanf("%d", &m);
    for (int i = 0; i < (m); i++) scanf(q[i]), g[i] = q[i];
    sort(g, g + m);
    int tot = unique(g, g + m) - g;
    map<int, long long> ans;
    map<int, int> dp[2];
    int now = 0, pre = 1;
    for (int i = (1); i <= (n); i++) {
      swap(now, pre);
      dp[now].clear();
      dp[now][a[i]] = 1;
      for (auto it : dp[pre]) dp[now][gcd(a[i], it.first)] += it.second;
      for (auto it : dp[now]) ans[it.first] += it.second;
    }
    for (int i = 0; i < (m); i++)
      if (ans.count(q[i]))
        printf("%I64d\n", ans[q[i]]);
      else
        puts("0");
  }
}