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 brp, a[333555], d[3000555], in[3000555], que[3000555];
vector<int> adj[3000555];
pair<int, int> par[3000555];
map<int, long long> res;
map<pair<int, int>, int> mapa;
int ubaci(int t, int i, int td) {
  pair<int, int> p = make_pair(t, i);
  brp++;
  d[brp] = td;
  in[brp] = 0;
  par[brp] = p;
  mapa[p] = brp;
  return brp;
}
int indeks(int t, int i) {
  pair<int, int> p = make_pair(t, i);
  if (mapa.count(p) > 0) {
    return mapa[p];
  } else {
    return -1;
  }
}
void updres(int x, long long d) {
  long long t;
  if (res.count(x) > 0) {
    t = res[x];
    res[x] = t + d;
  } else {
    res[x] = d;
  }
}
long long nadjires(int x) {
  if (res.count(x) > 0) {
    return res[x];
  } else {
    return 0;
  }
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, i, j, t, x, y, l, q, lp, np, kr, poc;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  brp = 0;
  for (i = n - 1; i >= 0; i--) {
    lp = ubaci(a[i], i, 1);
    t = a[i];
    for (j = i + 1; j < n; j++) {
      t = gcd(t, a[j]);
      np = indeks(t, j);
      if (np == -1) {
        np = ubaci(t, j, 0);
        adj[lp].push_back(np);
        in[np]++;
      } else {
        adj[lp].push_back(np);
        in[np]++;
        break;
      }
      lp = np;
    }
  }
  poc = 0;
  kr = 0;
  for (i = 1; i <= brp; i++) {
    if (in[i] == 0) {
      kr++;
      que[kr] = i;
    }
  }
  while (poc < kr) {
    poc++;
    x = que[poc];
    updres(par[x].first, d[x]);
    l = adj[x].size();
    for (i = 0; i < l; i++) {
      y = adj[x][i];
      d[y] += d[x];
      in[y]--;
      if (in[y] == 0) {
        kr++;
        que[kr] = y;
      }
    }
  }
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    scanf("%d", &x);
    printf("%I64d\n", nadjires(x));
  }
  return 0;
}