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 std ::unordered_map;
template <typename T>
inline void read(T &x) {
  x = 0;
  int fu = 1;
  char c = getchar();
  while (c > 57 || c < 48) {
    if (c == 45) fu = -1;
    c = getchar();
  }
  while (c <= 57 && c >= 48) {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  x *= fu;
}
template <typename T>
inline void fprint(T x) {
  if (x < 0) putchar(45), x = -x;
  if (x > 9) fprint(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void fprint(T x, char ch) {
  fprint(x);
  putchar(ch);
}
template <typename T, typename... Args>
inline void read(T &x, Args &...args) {
  read(x);
  read(args...);
}
inline int gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
int g[100005][21], lg[100005];
int n;
inline int query(int l, int r) {
  int k = lg[r - l + 1];
  return gcd(g[l][k], g[r - (1 << k) + 1][k]);
}
unordered_map<int, long long> cnt;
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(g[i][0]);
  lg[0] = -1;
  for (int i = 1; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
  for (int k = 1; k <= lg[n]; ++k)
    for (int i = 1; i + (1 << k) - 1 <= n; ++i)
      g[i][k] = gcd(g[i][k - 1], g[i + (1 << (k - 1))][k - 1]);
  for (int i = 1; i <= n; ++i) {
    int R = n;
    while (R >= i) {
      int key = query(i, R);
      int l = i, r = R, pos = i;
      while (l <= r) {
        int mid = (l + r) >> 1;
        if (query(i, mid) == key)
          r = mid - 1, pos = mid;
        else
          l = mid + 1;
      }
      cnt[key] += R - pos + 1;
      R = pos - 1;
    }
  }
  int q;
  read(q);
  while (q--) {
    int x;
    read(x);
    fprint(cnt[x], 10);
  }
}