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;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    char c = getchar();
    bool f = false;
    for (; !isdigit(c); c = getchar()) f |= c == '-';
    for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    if (f) x = -x;
  }
  template <typename T, typename... Arg>
  inline void read(T &x, Arg &...args) {
    read(x);
    read(args...);
  }
  template <typename T>
  inline void write(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  const int N = 1e6 + 100;
  int n, a[N];
  char c[N];
  int pr[N];
  vector<int> v[N];
  stack<int> s;
  signed main() {
    for (int i = 2; i < N; i++)
      if (!pr[i])
        for (int j = 1; i * j < N; j++) pr[i * j] = i;
    read(n);
    for (int i = 1; i <= n; i++) read(a[i]);
    fgets(c + 1, 998244353, stdin);
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      s.push(i);
      while (a[i] > 1) {
        int j = pr[a[i]];
        a[i] /= j;
        if (c[i] == '*')
          v[j].push_back(i);
        else {
          while (s.size() && (!v[j].size() || s.top() > v[j].back())) s.pop();
          if (v[j].size()) v[j].pop_back();
        }
      }
      ans += s.size();
    }
    write(ans);
  }
  