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;
int a[110000];
map<int, long long> m;
int gcd(int a, int b) { return a && b ? gcd(b, a % b) : a + b; }
void solve(int l, int r) {
  if (r - l <= 3) {
    for (int i = l; i <= r; i++) {
      int g = a[i];
      for (int j = i; j <= r; j++) {
        g = gcd(g, a[j]);
        m[g]++;
      }
    }
    return;
  }
  int mid = (l + r) >> 1;
  solve(l, mid);
  solve(mid + 1, r);
  vector<pair<int, long long> > ls;
  vector<pair<int, long long> > rs;
  int cur = -1;
  long long cnt = 0;
  int g = a[mid];
  for (int i = mid; i >= l; i--) {
    g = gcd(g, a[i]);
    if (g != cur) {
      if (cur != -1) ls.push_back(make_pair(cur, cnt));
      cur = g;
      cnt = 1;
    } else {
      ++cnt;
    }
  }
  ls.push_back(make_pair(cur, cnt));
  cur = -1;
  cnt = 0;
  g = a[mid + 1];
  for (int i = mid + 1; i <= r; i++) {
    g = gcd(g, a[i]);
    if (g != cur) {
      if (cur != -1) rs.push_back(make_pair(cur, cnt));
      cur = g;
      cnt = 1;
    } else {
      cnt++;
    }
  }
  rs.push_back(make_pair(cur, cnt));
  for (int i = 0; i < ls.size(); i++) {
    for (int j = 0; j < rs.size(); j++) {
      int g = gcd(ls[i].first, rs[j].first);
      long long num = ls[i].second * rs[j].second;
      m[g] += num;
    }
  }
}
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
    }
    m.clear();
    solve(1, n);
    int q, xi;
    scanf("%d", &q);
    while (q--) {
      scanf("%d", &xi);
      if (m.count(xi))
        printf("%I64d\n", m[xi]);
      else
        puts("0");
    }
  }
  return 0;
}