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;
  const int INF = 1e9 + 10, M = 1e6 + 137;
  int pr[M], a[M], till[M], res[M];
  string op;
  unordered_map<int, int> mp[M];
  void merge(int i, int j) {
    if (mp[i].size() < mp[j].size()) {
      mp[i].swap(mp[j]);
    }
    for (pair<long long, long long> u : mp[j]) mp[i][u.first] += u.second;
    mp[j].clear();
  }
  bool divide(int i, int j) {
    for (pair<long long, long long> u : mp[j]) {
      if (mp[i][u.first] < u.second) return 0;
    }
    for (pair<long long, long long> u : mp[j]) mp[i][u.first] -= u.second;
    return 1;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    fill(pr, pr + M, INF);
    for (int i = 2; i <= 1e6; i++) {
      if (pr[i] == INF) {
        for (int j = i; j <= 1e6; j += i) pr[j] = min(pr[j], i);
      }
    }
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      int x = a[i];
      while (x > 1) {
        mp[i][pr[x]]++;
        x /= pr[x];
      }
    }
    cin >> op;
    for (int i = 0; i < n; i++) {
      if (op[i] == '/' && a[i] == 1) op[i] = '*';
    }
    till[n] = n;
    long long ans = 0;
    for (int i = n - 1; ~i; i--) {
      if (op[i] == '*') {
        till[i] = i + 1;
        while (till[i] != n) {
          if (till[till[i]] != till[i]) {
            merge(i, till[i]);
            till[i] = till[till[i]];
          } else {
            if (divide(i, till[i])) {
              till[i] = till[i] + 1;
            } else {
              break;
            }
          }
        }
      } else {
        till[i] = i;
      }
      ans += till[i] - i;
    }
    cout << ans;
  }
  