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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long INF = 1e18 + 123;
  const long double EPS = 1e-9;
  const int inf = 1e9 + 123;
  const int MOD = 1e9 + 7;
  const int N = 2e5 + 123;
  const int M = 1e6 + 12;
  const double pi = 3.14159265359;
  const int dx[] = {0, 0, 1, -1};
  const int dy[] = {1, -1, 0, 0};
  int a[M];
  vector<pair<int, int> > v[M];
  bool bp[1000];
  void solve() {
    vector<int> pr;
    for (int i = 2; i < 1000; i++) {
      if (bp[i]) continue;
      pr.push_back(i);
      for (int j = i * i; j < 1000; j += i) {
        bp[j] = 1;
      }
    }
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    string s;
    cin >> s;
    long long ans = 0;
    set<int> bb = {inf};
    for (int i = n, mn; i >= 1; i--) {
      mn = n + 1;
      if (a[i] > 1 && s[i - 1] == '/') mn = i;
      for (int d : pr) {
        if (a[i] % d) {
          if (!v[d].empty()) {
            mn = min(mn, v[d].back().second);
          }
          continue;
        }
        int cnt = 0;
        while (a[i] % d == 0) {
          a[i] /= d;
          cnt++;
        }
        if (s[i - 1] == '*') {
          while (!v[d].empty()) {
            if (v[d].back().first <= cnt) {
              cnt -= v[d].back().first;
              v[d].pop_back();
            } else {
              mn = min(mn, v[d].back().second);
              (--v[d].end())->first -= cnt;
              break;
            }
          }
        } else {
          v[d].push_back({cnt, i});
        }
      }
      if (a[i] > 1) {
        int d = a[i];
        if (s[i - 1] == '*') {
          if (!v[d].empty()) {
            bb.erase(v[d].back().second);
            v[d].pop_back();
          }
        } else {
          v[d].push_back({1, i});
          bb.insert(i);
        }
      }
      mn = min(mn, *bb.begin());
      ans += mn - i;
    }
    cout << ans;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    for (int i = 1; i <= t; i++) {
      solve();
    }
    return 0;
  }
  