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;
using int64 = long long;
using vi = vector<int>;
using vi64 = vector<int64>;
using mii = map<int, int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
using vpii = vector<pii>;
const int inf = 2e9 + 517;
const int mod = 1e9 + 7;
const char nwl = '\n';
const char tab = '\t';
const char space = ' ';
const int mxx = 1e5 + 543;
int64 sumV(vi &v) {
  int64 s = 0;
  for (int x : v) {
    s += (int64)x;
  }
  return s;
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
bool powerOf2(int n) {
  if (n & (n - 1)) return 0;
  return 1;
}
int arr[mxx], dp[26][mxx];
int fun(int a, int b) { return gcd(a, b); }
void buildtable(int n) {
  for (int i = 0; i < n; i++) {
    dp[0][i] = arr[i];
  }
  int k = log2(n);
  for (int i = 1; i <= k; i++) {
    int p = pow(2, i - 1);
    for (int j = 0; j < n; j++) {
      if (j + p * 2 <= n) {
        dp[i][j] = fun(dp[i - 1][j], dp[i - 1][j + p]);
      } else {
        break;
      }
    }
  }
}
int getbw_l_r(int l, int r) {
  if (l == r) return arr[l];
  int p = (r - l + 1);
  int c = log2(p);
  p = 1 << c;
  return fun(dp[c][l], dp[c][r - p + 1]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  buildtable(n);
  map<int, int64> res;
  for (int i = 0; i < n; i++) {
    int cur = arr[i];
    int last = i;
    while (last < n) {
      int low = last;
      int up = n - 1;
      while (low < up) {
        int m = (up - low + 1) / 2;
        m += low;
        int u = getbw_l_r(i, m);
        if (u == cur)
          low = m;
        else
          up = m - 1;
      }
      res[cur] += (int64)(low - last + 1);
      last = low + 1;
      cur = getbw_l_r(i, last);
    }
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int x;
    cin >> x;
    cout << res[x] << nwl;
  }
  return 0;
}