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>
  using MinHeap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  using MaxHeap = priority_queue<T>;
  const long double PI = 3.14159265358979323846;
  const long double E = 2.718281828459;
  long long MOD = 1e9 + 7;
  long long M(long long n) { return (n % MOD + MOD) % MOD; }
  long long pow2(long long n) { return 1ll << n; }
  mt19937 rng(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  const long long MX = 1e6 + 2;
  const long long MAXN = MX;
  vector<int> prime;
  bool is_composite[MAXN];
  vector<int> minDiv(MAXN);
  void sieve(long long n) {
    std::fill(is_composite, is_composite + n, false);
    for (long long i = 2; i < n; ++i) {
      if (!is_composite[i]) {
        prime.push_back(i);
        minDiv[i] = i;
      }
      for (long long j = 0; j < prime.size() && i * prime[j] < n; ++j) {
        is_composite[i * prime[j]] = true;
        minDiv[i * prime[j]] = prime[j];
        if (i % prime[j] == 0) break;
      }
    }
  }
  unordered_map<int, stack<int>> suf;
  int main(int argn, char **argv) {
    ios::sync_with_stdio(false);
    cout.tie(nullptr);
    cin.tie(nullptr);
    sieve(MX);
    ;
    int T = 1;
    for (long long(I) = (0); (I) <= ((T)-1); (I)++) {
      long long n;
      cin >> n;
      vector<int> a(n);
      for (long long(i) = (0); (i) <= ((n)-1); (i)++) cin >> a[i];
      string s;
      cin >> s;
      long long ans = 0;
      set<int> lim;
      lim.emplace(n);
      for (long long(i) = (n - 1); (i) >= (0); (i)--) {
        if (s[i] == '/' && a[i] > 1) lim.emplace(i);
        long long tmp = a[i];
        while (tmp > 1) {
          long long pr = minDiv[tmp], c = 0;
          while (minDiv[tmp] == pr) {
            tmp /= pr, c++;
          }
          if (s[i] == '*') {
            while (!suf[pr].empty() && c) {
              auto idx = suf[pr].top();
              c--, a[idx] /= pr;
              if (a[idx] % pr) suf[pr].pop();
              if (a[idx] == 1) lim.erase(idx);
            }
          } else {
            suf[pr].emplace(i);
          }
        };
        long long maxPos = *lim.begin(), earn = maxPos - i;
        ;
        ans += earn;
      }
      cout << ans << endl;
    }
    return 0;
  }
  