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>
  #pragma GCC optimize(3, "Ofast", "inline")
  using namespace std;
  long long n, a[1000006], ret;
  char str[1000006];
  vector<long long> b[1000006], v[1000006];
  stack<long long> s;
  void pre() {
    for (long long i = 2; i < 1000006; i++) {
      if (b[i].empty()) {
        for (long long j = i; j < 1000006; j *= i) {
          for (long long k = j; k < 1000006; k += j) {
            b[k].push_back(i);
          }
        }
      }
    }
  }
  signed main() {
    pre();
    scanf("%lld", &n);
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    scanf("%s", str + 1);
    for (long long i = 1; i <= n; i++) {
      char c = str[i];
      s.push(i);
      for (long long j : b[a[i]]) {
        if (c == '*') {
          v[j].push_back(i);
        } else {
          while (!s.empty() && (v[j].empty() || s.top() > v[j].back())) {
            s.pop();
          }
          if (!v[j].empty()) v[j].pop_back();
        }
      }
      ret += s.size();
    }
    cout << ret << endl;
    return 0;
  }
  