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 N = 100000;
int a[N];
int n;
const int M = 2000;
int G[M][M];
int GCD(int a, int b) { return b > 0 ? GCD(b, a % b) : a; }
int gcd(int a, int b) {
  if (a < M && b < M) {
    return G[a][b];
  }
  return b > 0 ? gcd(b, a % b) : a;
}
struct S {
  vector<int> t;
  void build() {
    t.resize(2 * n);
    for (int i = 0; i < n; ++i) {
      t[i + n] = a[i];
    }
    for (int i = 2 * n - 1; i > 0; i -= 2) {
      t[i >> 1] = gcd(t[i], t[i ^ 1]);
    }
  }
  int get(int l, int r) {
    l += n;
    r += n;
    int res = 0;
    for (; l <= r && res != 1; l = (l + 1) >> 1, r = (r - 1) >> 1) {
      if (l & 1) {
        res = gcd(t[l], res);
      }
      if (!(r & 1)) {
        res = gcd(t[r], res);
      }
    }
    return res;
  }
};
S s;
unordered_map<int, long long> ans;
struct Q {
  pair<map<int, int>, map<int, int> > pa;
  int g;
};
Q build(int l, int r) {
  if (l == r) {
    ans[a[l]]++;
    map<int, int> res;
    res[a[l]]++;
    Q ret;
    ret.g = a[l];
    ret.pa = make_pair(res, res);
    return ret;
  }
  int m = (l + r) >> 1;
  Q pl = build(l, m);
  Q pr = build(m + 1, r);
  for (pair<int, int> p1 : pl.pa.second) {
    for (pair<int, int> p2 : pr.pa.first) {
      ans[gcd(p1.first, p2.first)] += p1.second * 1ll * p2.second;
    }
  }
  Q res;
  res.g = gcd(pl.g, pr.g);
  res.pa.first = pl.pa.first;
  for (pair<int, int> p2 : pr.pa.first) {
    res.pa.first[gcd(p2.first, pl.g)] += p2.second;
  }
  res.pa.second = pr.pa.second;
  for (pair<int, int> p1 : pl.pa.second) {
    res.pa.second[gcd(p1.first, pr.g)] += p1.second;
  }
  return res;
}
int main() {
  for (int i = 0; i < M; ++i) {
    for (int j = i; j < M; ++j) {
      G[i][j] = G[j][i] = GCD(i, j);
    }
  }
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  build(0, n - 1);
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int x;
    cin >> x;
    cout << ans[x] << "\n";
  }
  return 0;
}