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 long long N = 5e5 + 5, Log = 30;
long long n, q, u, l, st[N][Log];
map<long long, long long> m;
inline long long gcd(long long x, long long y) {
  return (y ? gcd(y, x % y) : x);
}
long long ck(long long lf, long long rt) {
  long long j = log2(rt - lf + 1);
  return gcd(st[lf][j], st[rt - (1 << j) + 1][j]);
}
long long bs(long long lf, long long rt, long long aim, long long s) {
  long long md = (lf + rt) >> 1, res = -1;
  while (lf <= rt) {
    if (ck(s, md) == aim)
      res = md, lf = md + 1;
    else
      rt = md - 1;
    md = (lf + rt) >> 1;
  }
  return res + 1;
}
long long read() {
  long long x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - 48, ch = getchar();
  return x;
}
signed main() {
  n = read();
  l = log2(n) + 1;
  for (long long i = 1; i <= n; i++) st[i][0] = read();
  for (long long j = 1; j <= l; j++)
    for (long long i = 1; i <= n; i++)
      st[i][j] = gcd(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  for (long long i = 1; i <= n; i++) {
    long long now = i, lst, g;
    while (now <= n) {
      lst = now;
      g = ck(i, now);
      now = bs(now, n, g, i);
      m[g] += now - lst;
    }
  }
  q = read();
  while (q--) {
    u = read();
    printf("%lld\n", m[u]);
  }
  return 0;
}