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 long long N = 1000009;
  const long long INF = 0x3f3f3f3f;
  long long dta[N];
  long long L[N];
  char sing[N];
  long long prime[N];
  vector<long long> pos[N];
  signed main(void) {
    for (long long i = 2; i < N; i++)
      if (prime[i] == 0) {
        prime[i] = i;
        for (long long j = i * i; j < N; j += i) prime[j] = i;
      }
    long long n;
    long long temp;
    cin >> n;
    for (long long i = 1; i <= n; i++) L[i] = INF;
    for (long long i = 1; i <= n; i++) cin >> dta[i];
    cin >> sing + 1;
    for (long long i = 1; i <= n; i++) {
      L[i] = i;
      if (sing[i] == '*') {
        temp = dta[i];
        while (temp > 1) {
          pos[prime[temp]].push_back(i);
          temp /= prime[temp];
        }
      } else {
        temp = dta[i];
        while (temp > 1) {
          if (pos[prime[temp]].empty()) {
            L[i] = -1;
            break;
          }
          L[i] = min(L[i], pos[prime[temp]].back());
          pos[prime[temp]].pop_back();
          temp /= prime[temp];
        }
      }
    }
    long long ans = 0;
    vector<long long> vec;
    for (long long i = 1; i <= n; i++) {
      while (!vec.empty() && vec.back() > L[i]) vec.pop_back();
      if (i == L[i]) vec.push_back(i);
      ans += vec.size();
    }
    cout << ans << endl;
    return 0;
  }
  