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;
template <typename T>
inline void read(T &x) {
  T f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
const int maxn = 300010;
int n, m, x, a[maxn];
map<int, long long> ans;
vector<pair<int, long long> > ql, qr;
int gcd(int a, int b) { return a && b ? gcd(b, a % b) : a + b; }
void Solve(int l, int r) {
  if (r - l <= 3) {
    for (int i = l, d; i <= r; i++) {
      d = a[i];
      for (int j = i; j <= r; j++) {
        d = gcd(d, a[j]);
        ans[d]++;
      }
    }
    return;
  }
  int mid = (l + r) >> 1, d, lst = -1;
  Solve(l, mid);
  Solve(mid + 1, r);
  ql.clear();
  qr.clear();
  long long cnt = 0;
  d = a[mid];
  for (int i = mid; i >= l; i--) {
    d = gcd(d, a[i]);
    if (d != lst) {
      if (lst != -1) ql.push_back(make_pair(lst, cnt));
      lst = d;
      cnt = 1;
    } else
      cnt++;
  }
  ql.push_back(make_pair(d, cnt));
  lst = -1;
  cnt = 0;
  d = a[mid + 1];
  for (int i = mid + 1; i <= r; i++) {
    d = gcd(d, a[i]);
    if (d != lst) {
      if (lst != -1) qr.push_back(make_pair(lst, cnt));
      lst = d;
      cnt = 1;
    } else
      cnt++;
  }
  qr.push_back(make_pair(d, cnt));
  for (auto i = ql.begin(); i != ql.end(); i++)
    for (auto j = qr.begin(); j != qr.end(); j++)
      ans[gcd(i->first, j->first)] += i->second * j->second;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  ans.clear();
  Solve(1, n);
  read(m);
  while (m--) {
    read(x);
    if (ans.count(x))
      printf("%lld\n", ans[x]);
    else
      puts("0");
  }
  return 0;
}