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;
  set<pair<int, int> > s[1000005];
  int bio[1000005], r[1000005], arr[1000005];
  vector<pair<int, int> > veki[1000005];
  char b[1000005];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    for (int i = 2; i <= 1000000; i++) {
      if (!bio[i]) {
        for (int j = i; j <= 1000000; j += i) {
          bio[j] = 1;
          int v = j;
          int br = 0;
          while (v % i == 0) {
            v /= i;
            br++;
          }
          veki[j].push_back(make_pair(i, br));
        }
      }
    }
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    for (int i = 1; i <= n; i++) {
      cin >> b[i];
    }
    long long sol = 0ll;
    for (int i = n; i > 0; i--) {
      if (b[i] == '/' and arr[i] > 1) {
        r[i] = i;
        continue;
      }
      if (b[i] == '*') {
        for (auto v : veki[arr[i]]) {
          s[i].insert(v);
        }
      }
      int j = i + 1;
      r[i] = i + 1;
      while (j <= n) {
        if (r[j] > j) {
          if (s[j].size() > s[i].size()) swap(s[i], s[j]);
          while (s[j].size()) {
            pair<int, int> p = *s[j].begin();
            s[j].erase(s[j].begin());
            auto it = s[i].lower_bound(make_pair(p.first, 0));
            if (it == s[i].end() or (*it).first > p.first)
              s[i].insert(p);
            else {
              p.second += (*it).second;
              s[i].erase(it);
              s[i].insert(p);
            }
          }
          j = r[j];
        } else {
          int tr = 1;
          for (auto v : veki[arr[j]]) {
            auto it = s[i].lower_bound(make_pair(v.first, 0));
            if (it == s[i].end() or (*it).first > v.first) {
              tr = 0;
              break;
            }
            if ((*it).second < v.second) {
              tr = 0;
              break;
            }
          }
          if (!tr) {
            break;
          } else {
            for (auto v : veki[arr[j]]) {
              auto it = s[i].lower_bound(make_pair(v.first, 0));
              pair<int, int> p = make_pair(v.first, (*it).second - v.second);
              s[i].erase(it);
              s[i].insert(p);
            }
            j++;
          }
        }
      }
      r[i] = j;
    }
    for (int i = 1; i <= n; i++) {
      sol += r[i] - i;
    }
    cout << sol;
    return 0;
  }
  