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;
  void solve();
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
  }
  const long long N = 1e6 + 100, A = 1e6 + 100, K = 1001, INF = 2e18,
                  MOD = 1e9 + 7;
  long long n, a[N], g[N], prime[N];
  vector<long long> v[N];
  string s;
  vector<long long> primes(long long e) {
    vector<long long> res;
    while (e > 1) {
      res.push_back(prime[e]);
      e /= prime[e];
    }
    return res;
  }
  void solve() {
    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;
        }
      }
    }
    cin >> n;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    cin >> s;
    for (long long i = 0; i < n; i++) {
      vector<long long> pr = primes(a[i]);
      if (s[i] == '*') {
        g[i] = i;
        for (long long x : pr) {
          v[x].push_back(i);
        }
      } else {
        long long mn = INF;
        for (long long x : pr) {
          if (v[x].empty()) {
            mn = -1;
          } else {
            mn = min(mn, v[x].back());
            v[x].pop_back();
          }
        }
        g[i] = mn;
      }
    }
    long long ans = 0;
    vector<pair<long long, long long>> u;
    for (long long i = n - 1; i >= 0; i--) {
      u.push_back({g[i], i});
      while (!u.empty() && u.back().first >= i) {
        u.pop_back();
      }
      if (u.empty()) {
        ans += n - i;
      } else {
        ans += u.back().second - i;
      }
    }
    cout << ans;
  }
  