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 read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int bin[20], Log[100005];
int n, q;
int a[100005], x[300005], f[100005][17];
map<int, long long> ans;
struct data {
  int l, r, v;
} t[400005];
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
void pre() {
  for (int i = 1; i <= n; i++) f[i][0] = a[i];
  for (int i = 1; i <= 16; i++)
    for (int x = 1; x <= n; x++)
      if (x + bin[i] - 1 <= n)
        f[x][i] = gcd(f[x][i - 1], f[x + bin[i - 1]][i - 1]);
      else
        break;
}
int query(int l, int r) {
  int t = Log[r - l + 1];
  return gcd(f[l][t], f[r - bin[t] + 1][t]);
}
int find(int x, int l, int L) {
  int r = n, ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (query(L, mid) != x)
      r = mid - 1;
    else
      ans = mid, l = mid + 1;
  }
  return ans;
}
void solve(int x) {
  int t = a[x], now = x;
  while (1) {
    int last = now;
    now = find(t, now, x);
    if (ans[t]) ans[t] += now - last + 1;
    if (now == n) return;
    now++;
    t = query(x, now);
  }
}
int main() {
  bin[0] = 1;
  for (int i = 1; i < 20; i++) bin[i] = bin[i - 1] << 1;
  n = read();
  Log[0] = -1;
  for (int i = 1; i <= n; i++) Log[i] = Log[i >> 1] + 1;
  for (int i = 1; i <= n; i++) a[i] = read();
  pre();
  q = read();
  for (int i = 1; i <= q; i++) x[i] = read(), ans[x[i]] = 1;
  for (int i = 1; i <= n; i++) solve(i);
  for (int i = 1; i <= q; i++) printf("%I64d\n", ans[x[i]] - 1);
  return 0;
}