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 = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int gcd(int a, int b) {
  if (b % a == 0) return a;
  return gcd(b % a, a);
}
const int Size = 100005;
long long n, a[Size];
map<int, long long> res;
long long stgcd[Size][21];
void init() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) stgcd[i][0] = a[i];
  for (int j = 1; j < 20; j++) {
    for (int i = 1 << j; i <= n; i++) {
      stgcd[i][j] = gcd(stgcd[i][j - 1], stgcd[i - (1 << (j - 1))][j - 1]);
    }
  }
}
int Query(int l, int r) {
  int x = log2(r - l + 1);
  return gcd(stgcd[l + (1 << x) - 1][x], stgcd[r][x]);
}
int Binary(int l, int r, int pos, int k) {
  int ans = r + 1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (Query(pos, mid) == k) {
      l = mid + 1;
    } else {
      r = mid - 1;
      if (mid < ans) ans = mid;
    }
  }
  if (ans == r) {
    return r;
  } else {
    return ans - 1;
  }
}
void work() {
  int ans, now, nx;
  for (int i = 1; i <= n; i++) {
    now = a[i];
    nx = i;
    while (1) {
      int tmp = nx;
      nx = Binary(tmp, n, i, now);
      now = Query(i, nx);
      res[now] += nx - tmp + 1;
      if (nx < n) {
        nx++;
        now = Query(i, nx);
      } else {
        break;
      }
    }
  }
}
int main() {
  init();
  work();
  int q = read(), k;
  while (q--) {
    k = read();
    printf("%I64d\n", res[k]);
  }
  return 0;
}