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;
  int N, a[1000005], b[1000005];
  string ss;
  bool IsPrime[1000006];
  int mp[1000006], Index, Prime[1000006], L[1000006];
  vector<int> stk[1000006];
  const int mx = 1e6 + 5;
  void Pre() {
    for (int i = 2; i <= mx; i++) {
      if (!IsPrime[i]) {
        mp[i] = i;
        Prime[++Index] = i;
      }
      for (int j = 1; j <= Index && i * Prime[j] <= mx; j++) {
        IsPrime[i * Prime[j]] = true;
        mp[i * Prime[j]] = Prime[j];
        if (i % Prime[j] == 0) break;
      }
    }
  }
  void Add(int x, int pos) {
    L[pos] = pos;
    while (x > 1) {
      stk[mp[x]].push_back(pos);
      x /= mp[x];
    }
  }
  void Del(int x, int pos) {
    if (x == 1) {
      L[pos] = pos;
      return;
    }
    L[pos] = pos;
    while (x > 1) {
      if (stk[mp[x]].empty()) {
        L[pos] = 0;
        return;
      }
      L[pos] = min(L[pos], stk[mp[x]].back());
      stk[mp[x]].pop_back();
      x /= mp[x];
    }
  }
  int main() {
    Pre();
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%d", &a[i]);
    cin >> ss;
    for (int i = 0; i < ss.length(); i++) {
      if (ss[i] == '*')
        Add(a[i + 1], i + 1);
      else
        Del(a[i + 1], i + 1);
    }
    long long ans = 0;
    vector<pair<int, int> > ff;
    for (int i = N; i >= 1; i--) {
      int cnt = 1;
      while (ff.size() > 0 && ff.back().first >= L[i]) {
        cnt += ff.back().second;
        ff.pop_back();
      }
      ff.push_back(make_pair(L[i], cnt));
      if (L[i] == i) ans += (long long)cnt;
    }
    printf("%lld", ans);
    return 0;
  }
  