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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
using namespace std;
template <typename T>
void uin(T &a, T b) {
  if (b < a) {
    a = b;
  }
}
template <typename T>
void uax(T &a, T b) {
  if (b > a) {
    a = b;
  }
}
const long long N = 100 * 1000 + 228;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long a[N];
long long st[N][20], lg[N];
long long get(long long l, long long r) {
  return gcd(st[l][lg[r - l + 1]],
             st[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  map<long long, long long> cnt;
  for (long long i = 1; i <= n; ++i) st[i][0] = a[i];
  lg[1] = 0;
  for (long long i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
  for (long long k = 1; k < 20; ++k) {
    for (long long i = 1; i <= n - (1 << k) + 1; ++i) {
      st[i][k] = gcd(st[i][k - 1], st[i + (1 << (k - 1))][k - 1]);
    }
  }
  for (long long i = 1; i <= n; ++i) {
    long long j = i;
    while (j <= n) {
      long long x = get(i, j);
      long long l = j, r = n + 1;
      while (r - l > 1) {
        long long mid = (l + r) >> 1;
        if (get(i, mid) == x)
          l = mid;
        else
          r = mid;
      }
      cnt[x] += l - j + 1;
      j = l + 1;
    }
  }
  long long q;
  cin >> q;
  while (q--) {
    long long x;
    cin >> x;
    cout << cnt[x] << '\n';
  }
  return 0;
}