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;
const int MOD = 1e8 + 7, inv2 = (MOD + 1) / 2;
double eps = 1e-6;
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
struct node {
  int x, y, z;
} b[100010], c[100010];
map<int, long long> ans;
int a[100010];
int read() {
  int v = 0, f = 1;
  char c = getchar();
  while (c < 48 || 57 < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (48 <= c && c <= 57) v = (v << 3) + v + v + c - 48, c = getchar();
  return v * f;
}
long long readll() {
  long long v = 0, f = 1;
  char c = getchar();
  while (c < 48 || 57 < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (48 <= c && c <= 57) v = (v << 3) + v + v + c - 48, c = getchar();
  return v * f;
}
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  int n = read();
  for (int i = (int)1; i <= (int)n; i++) a[i] = read();
  int m = 0;
  for (int i = (int)1; i <= (int)n; i++) {
    for (int j = (int)1; j <= (int)m; j++) b[j].z = gcd(b[j].z, a[i]);
    m++;
    b[m].x = b[m].y = i;
    b[m].z = a[i];
    int t = 0;
    for (int j = (int)1; j <= (int)m; j++)
      if (b[j].z != c[t].z)
        t++, c[t] = b[j];
      else
        c[t].y = b[j].y;
    m = t;
    for (int j = (int)1; j <= (int)m; j++) {
      b[j] = c[j];
      ans[b[j].z] = ans[b[j].z] + b[j].y - b[j].x + 1;
    }
  }
  int q = read();
  for (int i = (int)1; i <= (int)q; i++) {
    int x = read();
    printf("%I64d\n", ans[x]);
  }
  return 0;
}