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;
  vector<int> prime;
  void build(int n) {
    prime = vector<int>(n + 1);
    iota(prime.begin(), prime.end(), 0);
    for (int i = 2; i * i <= n; i++) {
      if (prime[i] == i) {
        for (int j = i * i; j <= n; j += i) {
          if (prime[j] == j) {
            prime[j] = i;
          }
        }
      }
    }
  }
  vector<int> factorize(int n) {
    vector<int> prime_div;
    while (n > 1) {
      int curr = prime[n];
      while (prime[n] == curr) {
        prime_div.push_back(curr);
        n /= curr;
      }
    }
    return prime_div;
  }
  void foo() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    build(*max_element(a.begin(), a.end()) + 5);
    string s;
    cin >> s;
    stack<int> st;
    vector<unordered_map<int, int>> st2;
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] == 1) {
        st.push(i);
        st2.emplace_back();
        continue;
      }
      if (s[i] == '*') {
        st.push(i);
        st2.emplace_back();
      }
      auto divide = [&](int p) {
        while (st.size()) {
          int &v = st2.back()[p];
          if (v == 0) {
            ans += i - st.top();
            if (st2.size() > 1) {
              auto &pm = st2[st2.size() - 2];
              if (st2.back().size() > pm.size()) {
                pm.swap(st2.back());
              }
              for (const auto &[k, val] : st2.back()) {
                pm[k] += val;
              }
            }
            st.pop();
            st2.pop_back();
          } else {
            v--;
            break;
          }
        }
      };
      const vector<int> f = factorize(a[i]);
      if (s[i] == '*') {
        for (int p : f) {
          st2.back()[p]++;
        }
      } else {
        for (int p : f) {
          divide(p);
        }
      }
    }
    while (st.size()) {
      ans += n - st.top();
      st.pop();
    }
    cout << ans << endl;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    foo();
  }
  