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 = 100010;
class Q {
 public:
  int gcd;
  long long tot;
} c[5000000], d[5000000];
int cmp(const Q &a, const Q &b) { return a.gcd < b.gcd; }
vector<pair<pair<int, int>, int> > a[N];
int n, q, b[N], tot, total, x;
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
  a[n].push_back(make_pair(make_pair(n, n), b[n]));
  c[++tot].gcd = b[n];
  c[tot].tot = 1;
  for (int i = n - 1; i >= 1; --i) {
    int l = i, r = i, tmp = b[i];
    for (int j = 0; j < a[i + 1].size(); ++j) {
      if (tmp != gcd(tmp, a[i + 1][j].second)) {
        a[i].push_back(make_pair(make_pair(l, r), tmp));
        c[++tot].gcd = tmp;
        c[tot].tot = r - l + 1;
        l = a[i + 1][j].first.first, r = a[i + 1][j].first.second;
        tmp = gcd(tmp, a[i + 1][j].second);
      }
      r = a[i + 1][j].first.second;
      if (j == a[i + 1].size() - 1) {
        a[i].push_back(make_pair(make_pair(l, r), tmp));
        c[++tot].gcd = tmp;
        c[tot].tot = r - l + 1;
      }
    }
  }
  std::sort(c + 1, c + tot + 1, cmp);
  for (int i = 1; i <= tot; ++i)
    if (i == 1 || c[i].gcd != c[i - 1].gcd)
      d[++total].gcd = c[i].gcd, d[total].tot = c[i].tot;
    else
      d[total].tot += c[i].tot;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d", &x);
    int l = 1, r = total + 1;
    for (; l + 1 < r;) {
      int mid = (l + r) / 2;
      if (d[mid].gcd <= x)
        l = mid;
      else
        r = mid;
    }
    bool succ = false;
    for (int i = max(1, l - 3); i <= min(total, l + 3); ++i)
      if (d[i].gcd == x) {
        printf("%I64d\n", d[i].tot);
        succ = true;
        break;
      }
    if (!succ) printf("0\n");
  }
  return 0;
}