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;
  string _reset = "\u001b[0m", _yellow = "\u001b[33m", _bold = "\u001b[1m";
  void DBG() { cerr << "]" << _reset << endl; }
  template <class H, class... T>
  void DBG(H h, T... t) {
    cerr << to_string(h);
    if (sizeof...(t)) cerr << ", ";
    DBG(t...);
  }
  const int INF = 1e9 + 7, N = 1e6 + 7;
  int a[N], lp[N];
  int n;
  multiset<int> ms;
  struct DS {
    vector<pair<int, int> > tt;
    int lz, ans;
    DS() { lz = 0, ans = -1; }
    void undo() {
      if (ans != -1) ms.erase(ms.find(ans));
    }
    void godo() {
      if (tt.empty())
        ans = n;
      else
        ans = tt.back().first;
      ms.insert(ans);
    }
    void upd(int pos, int op) {
      undo();
      tt.emplace_back(pos, -lz);
      lz += op;
      while (tt.size() && tt.back().second + lz >= 0) tt.pop_back();
      godo();
    }
  } ds[N];
  long long ans;
  signed main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);
    ;
    for (int i = 2; i < N; i++) {
      if (lp[i]) continue;
      lp[i] = i;
      for (int j = 2 * i; j < N; j += i) lp[j] = i;
    }
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    string s;
    cin >> s;
    ans = 0;
    ms.insert(n);
    auto upd = [&](int x, int pos, int op) {
      while (x != 1) {
        int p = lp[x];
        int p_cnt = 0;
        while (x % p == 0) {
          p_cnt++;
          x /= lp[x];
        }
        ds[p].upd(pos, op * p_cnt);
      }
    };
    for (int i = n - 1; i >= 0; i--) {
      if (a[i] == 1) s[i] = '*';
      if (s[i] == '*')
        upd(a[i], i, 1);
      else
        upd(a[i], i, -1);
      ans += *ms.begin() - i;
    }
    cout << ans << '\n';
  }
  