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>
  #pragma GCC optimize("O3")
  using namespace std;
  int t, n, q;
  int v[1000010];
  string s;
  vector<int> pd[1000010];
  vector<int> cod[1000010];
  int p[1000010], siz;
  set<int> ses;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    for (int i = 2; i < 1000010; i++) {
      if (!p[i]) {
        for (int j = i; j < 1000010; j += i) {
          p[j] = 1;
          pd[j].push_back(i);
        }
      }
    }
    cin >> n;
    for (int i = 0; i < n; i++) cin >> v[i];
    cin >> s;
    long long ris = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] == '/') {
        for (auto j : pd[v[i]]) {
          cod[j].push_back(i);
          ses.insert(i);
        }
        if (v[i] == 1) {
          int mim = n;
          if (ses.size()) mim = *ses.begin();
          ris += mim - i;
        }
      } else {
        int kek = v[i], mim = n;
        for (auto j : pd[v[i]]) {
          while (kek % j == 0 && cod[j].size()) {
            int k = cod[j].back();
            v[k] /= j;
            if (v[k] == 1) {
              ses.erase(ses.find(k));
            }
            if (v[k] % j) {
              cod[j].pop_back();
            }
            kek /= j;
          }
        }
        if (ses.size()) mim = *ses.begin();
        ris += mim - i;
      }
    }
    cout << ris << "\n";
  }
  