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 gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
map<int, long long> ans;
int n, q, a[100005], st[100005][21], que[100005 << 2], Log[100005];
void init() {
  Log[0] = -1;
  for (int i = 1; i <= n; i++) Log[i] = Log[i >> 1] + 1;
  for (int i = 1; i <= n; i++) st[i][0] = a[i];
  for (int j = 1; j < 20; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      st[i][j] = gcd(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
}
int ask(int l, int r) {
  int ll = Log[r - l + 1];
  return gcd(st[l][ll], st[r - (1 << ll) + 1][ll]);
}
int erfen(int L, int l, int g) {
  int r = n, pos = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (ask(L, mid) == g) {
      pos = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return pos;
}
void solve(int l) {
  int la, r = l, t = a[l];
  while (1) {
    la = r;
    r = erfen(l, r, t);
    if (ans[t]) ans[t] += r - la + 1;
    if (r == n) return;
    r++;
    t = ask(l, r);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%d", &que[i]);
    ans[que[i]] = 1;
  }
  init();
  for (int i = 1; i <= n; i++) solve(i);
  for (int i = 1; i <= q; i++) printf("%lld\n", ans[que[i]] - 1);
}