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;
  using LL = long long;
  constexpr int maxn = 1000000 + 1;
  vector<pair<int, int>> vp[maxn], ch[maxn];
  int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    for (int i = 2; i < maxn; i += 1)
      if (vp[i].empty())
        for (int j = i; j < maxn; j += i) {
          vp[j].emplace_back(i, 0);
          for (int k = j; k % i == 0; k /= i) vp[j].back().second += 1;
        }
    int n;
    cin >> n;
    vector<int> a(n);
    for (int& x : a) cin >> x;
    string s;
    cin >> s;
    vector<vector<int>> L(n), R(n);
    for (int i = 0; i < n; i += 1)
      for (auto [p, c] : vp[a[i]]) ch[p].emplace_back(i, s[i] == '*' ? c : -c);
    for (int i = 2; i < maxn; i += 1)
      if (ch[i].size()) {
        int m = ch[i].size();
        vector<int> sum(m);
        for (int j = 0; j < m; j += 1) {
          sum[j] = ch[i][j].second;
          if (j) sum[j] += sum[j - 1];
        }
        vector<pair<int, int>> st = {{sum[m - 1], m - 1}};
        for (int j = m - 1; j >= 0; j -= 1) {
          int p = j ? sum[j - 1] : 0;
          while (not st.empty() and st.back().first >= p) st.pop_back();
          st.emplace_back(p, j - 1);
          int tl = (int)st.size() - 2;
          if (tl != -1) {
            int l = j ? ch[i][j - 1].first + 1 : 0;
            int r = ch[i][j].first;
            int x = ch[i][st[tl].second].first - 1;
            L[l].push_back(x);
            R[r].push_back(x);
          }
        }
      }
    LL ans = 0;
    multiset<int> ms;
    for (int i = 0; i < n; i += 1) {
      for (int l : L[i]) ms.insert(l);
      if (ms.empty())
        ans += n - i;
      else
        ans += *ms.begin() - i + 1;
      for (int r : R[i]) ms.erase(ms.find(r));
    }
    cout << ans << "\n";
    return 0;
  }
  