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;
  vector<pair<int, int>> fac[1000005];
  bool comp[1000005];
  int n;
  int arr[1000005];
  vector<pair<int, int>> pos[1000005];
  string s;
  map<int, int> good[1000005];
  long long ans = 0;
  vector<int> p;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    for (int i = 2; i <= 1000000; i++) {
      if (comp[i]) {
        continue;
      }
      for (int j = i; j <= 1000000; j += i) {
        comp[j] = true;
        int x = j, cnt = 0;
        while (x % i == 0) {
          x /= i;
          cnt++;
        }
        fac[j].push_back({i, cnt});
      }
    }
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
      for (pair<int, int> j : fac[arr[i]]) {
        pos[j.first].push_back({i, j.second});
      }
    }
    cin >> s;
    for (int i = 2; i <= 1000000; i++) {
      vector<int> k = {0};
      for (int j = 0; j < pos[i].size(); j++) {
        int x = pos[i][j].first;
        if (s[x - 1] == '*') {
          for (int it = 1; it <= pos[i][j].second; it++) {
            k.push_back(x);
          }
        } else {
          for (int it = 2; it <= pos[i][j].second and k.size() > 1; it++) {
            k.pop_back();
          }
          good[i][x] = k[k.size() - 1];
          if (k.size() > 1) {
            k.pop_back();
          }
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      p.push_back(i);
      if (s[i - 1] == '/') {
        int x = 2e9;
        for (pair<int, int> j : fac[arr[i]]) {
          x = min(x, good[j.first][i]);
        }
        while (p.size() > 0 and p[p.size() - 1] > x) {
          p.pop_back();
        }
      }
      long long siz = p.size();
      ans += siz;
    }
    cout << ans << "\n";
    return 0;
  }
  