description: Kuzya started going to school. He was given math homework in which he was given an array a of length n and an array of symbols b of length n, consisting of symbols '*' and '/'.
  
  Let's denote a path of calculations for a segment [l; r] (1 ≤ l ≤ r ≤ n) in the following way: 
  
    * Let x=1 initially. For every i from l to r we will consequently do the following: if b_i= '*', x=x*a_i, and if b_i= '/', then x=(x)/(a_i). Let's call a path of calculations for the segment [l; r] a list of all x that we got during the calculations (the number of them is exactly r - l + 1). 
  
  
  
  For example, let a=[7, 12, 3, 5, 4, 10, 9], b=[/, *, /, /, /, *, *], l=2, r=6, then the path of calculations for that segment is [12, 4, 0.8, 0.2, 2].
  
  Let's call a segment [l;r] simple if the path of calculations for it contains only integer numbers. 
  
  Kuzya needs to find the number of simple segments [l;r] (1 ≤ l ≤ r ≤ n). Since he obviously has no time and no interest to do the calculations for each option, he asked you to write a program to get to find that number!
  
  Input
  
  The first line contains a single integer n (2 ≤ n ≤ 10^6).
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
  
  The third line contains n symbols without spaces between them — the array b_1, b_2 … b_n (b_i= '/' or b_i= '*' for every 1 ≤ i ≤ n).
  
  Output
  
  Print a single integer — the number of simple segments [l;r].
  
  Examples
  
  Input
  
  
  3
  1 2 3
  */*
  
  
  Output
  
  
  2
  
  
  Input
  
  
  7
  6 4 10 1 2 15 1
  */*/*//
  
  
  Output
  
  
  8
solution: #include <bits/stdc++.h>
  using namespace std;
  int T, n, p[1100000], l[1100000], a[1100000], ans[1100000];
  vector<int> v[1100000];
  string s;
  int main() {
    for (int i = 2; i <= 1000000; i++) {
      if (!p[i]) {
        p[i] = i;
        for (int j = 2; i * j <= 1000000; j++) {
          if (!p[i * j]) p[i * j] = i;
        }
      }
    }
    ios::sync_with_stdio(0);
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    cin >> s;
    for (int i = 1; i <= n; i++) {
      if (s[i - 1] == '*') {
        l[i] = i;
        while (a[i] != 1) {
          v[p[a[i]]].push_back(i);
          a[i] /= p[a[i]];
        }
      } else {
        l[i] = i;
        while (a[i] != 1) {
          if (v[p[a[i]]].empty())
            l[i] = -1;
          else
            l[i] = min(l[i], v[p[a[i]]].back()), v[p[a[i]]].pop_back();
          a[i] /= p[a[i]];
        }
      }
    }
    stack<int> s;
    long long sum = 0;
    for (int i = n; i >= 1; i--) {
      ans[i] = 1;
      while (!s.empty() && s.top() != -1 && l[s.top()] >= l[i])
        ans[i] += ans[s.top()], s.pop();
      s.push(i);
      if (l[i] == i) sum += ans[i];
    }
    cout << sum << endl;
    return 0;
  }
  