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>
  int n, a[1000005], from[1000005];
  long long ans;
  char b[1000005];
  std::vector<int> p;
  std::vector<std::pair<int, int>> app[1000005], sta, w[1000005];
  signed main() {
    std::ios::sync_with_stdio(false);
    std::cin >> n;
    for (int i = 1; i <= n; ++i) std::cin >> a[i];
    for (int i = 1; i <= n; ++i) std::cin >> b[i];
    for (int i = 2; i <= 1000000; ++i) {
      if (!from[i]) {
        from[i] = i;
        p.push_back(i);
      }
      for (auto j : p) {
        if (i * j > 1000000) break;
        from[i * j] = j;
        if (i % j == 0) break;
      }
    }
    for (int i = 1; i <= n; ++i)
      while (a[i] > 1) {
        int v = from[a[i]], tim = 0;
        while (a[i] % v == 0) a[i] /= v, ++tim;
        if (app[v].empty()) app[v].emplace_back(0, 0);
        app[v].push_back({(b[i] == '/' ? -1 : +1) * tim, i});
      }
    for (int i = 1; i <= 1000000; ++i) {
      if (app[i].empty()) continue;
      for (int j = 1; j != (int)app[i].size(); ++j)
        app[i][j].first += app[i][j - 1].first;
      app[i].push_back({INT_MIN, n + 1});
      sta.clear();
      sta.emplace_back(INT_MIN, n + 1);
      for (int j = app[i].size() - 1; j--;) {
        while (sta.back().first >= app[i][j].first) sta.pop_back();
        w[app[i][j + 1].second].emplace_back(sta.back().second, app[i][j].second);
        sta.emplace_back(app[i][j]);
      }
    }
    std::priority_queue<std::pair<int, int>, std::vector<std::pair<int, int>>,
                        std::greater<std::pair<int, int>>>
        que;
    que.push({n + 1, 0});
    for (int i = n; i; --i) {
      for (auto j : w[i]) que.push(j);
      while (que.top().second >= i) que.pop();
      ans += que.top().first - i;
    }
    std::cout << ans << std::endl;
    return 0;
  }
  