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;
  int const maxn = 1e6 + 5;
  int a[maxn];
  int prime[maxn];
  int L[maxn];
  vector<int> pos[maxn];
  inline void add(int x, int l) {
    L[l] = l;
    while (x > 1) {
      pos[prime[x]].push_back(l);
      x /= prime[x];
    }
  }
  inline void del(int x, int l) {
    if (x == 1) {
      L[l] = l;
      return;
    }
    L[l] = l;
    while (x > 1) {
      if ((int)pos[prime[x]].size() == 0) {
        L[l] = 0;
        return;
      }
      L[l] = min(L[l], pos[prime[x]].back());
      pos[prime[x]].pop_back();
      x /= prime[x];
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 2; i < maxn; ++i) {
      if (prime[i] == 0) {
        prime[i] = i;
        if (i > 1000) continue;
        for (int j = i * i; j < maxn; j += i) {
          prime[j] = i;
        }
      }
    }
    char type;
    for (int i = 1; i <= n; ++i) {
      cin >> type;
      if (type == '*')
        add(a[i], i);
      else
        del(a[i], i);
    }
    long long answer = 0;
    vector<pair<int, int> > f_min;
    for (int i = n; i >= 1; --i) {
      int cnt = 1;
      while ((int)f_min.size() > 0 && f_min.back().first >= L[i]) {
        cnt += f_min.back().second;
        f_min.pop_back();
      }
      f_min.push_back({L[i], cnt});
      if (L[i] == i) answer += (long long)cnt;
    }
    cout << answer << '\n';
    return 0;
  }
  