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<pair<int, int>> factors[1000005];
  vector<pair<int, int>> v[1000005];
  vector<int> nex[1000005];
  int arr[1000005];
  char brr[1000005];
  int sum[1000005];
  int ans[1000005];
  int pos[1000005];
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < 1000005; i++) ans[i] = n;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    for (int i = 1; i <= n; i++) {
      cin >> brr[i];
    }
    for (int i = 1; i <= n; i++) {
      for (auto u : factors[arr[i]]) {
        if (brr[i] == '/') {
          sum[u.first] -= u.second;
        }
        if (brr[i] == '*') {
          sum[u.first] += u.second;
        }
        v[u.first].push_back({sum[u.first], i});
        if (sum[u.first] < 0) ans[u.first] = min(ans[u.first], i - 1);
      }
    }
    for (int i = 0; i < 1000005; i++) {
      if (v[i].empty()) continue;
      stack<int> st;
      st.push(v[i].size());
      v[i].push_back({-1e9, n + 1});
      for (int j = v[i].size() - 2; j >= 0; j--) {
        while (v[i][j] <= v[i][st.top()]) st.pop();
        nex[i].push_back(v[i][st.top()].second - 1);
        st.push(j);
      }
      reverse(nex[i].begin(), nex[i].end());
    }
    multiset<int> s;
    s.insert(n);
    for (int i = 0; i < 1000005; i++)
      if (v[i].size()) s.insert(ans[i]);
    long long res = *s.begin();
    for (int i = 2; i <= n; i++) {
      for (auto u : factors[arr[i - 1]]) {
        s.erase(s.find(ans[u.first]));
        ans[u.first] = nex[u.first][pos[u.first]];
        pos[u.first]++;
        s.insert(ans[u.first]);
      }
      res += max(0, *s.begin() - i + 1);
    }
    cout << res << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    for (int i = 2; i < 1000005; i++) {
      if (factors[i].empty()) {
        for (int j = i; j < 1000005; j += i) {
          int tmp = j;
          int cnt = 0;
          while (tmp / i * i == tmp) {
            tmp /= i;
            cnt++;
          }
          factors[j].push_back({i, cnt});
        }
      }
    }
    int t = 1;
    while (t--) {
      solve();
    }
  }
  