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 a[1000000 + 5];
  char b[1000000 + 5];
  int las[1000000 + 5];
  void get_prime() {
    int n = 1000000;
    for (int i = 2; i <= n; i++)
      if (!las[i]) {
        las[i] = i;
        if (i > 1000) continue;
        for (int j = i * i; j <= n; j += i) las[j] = i;
      }
  }
  int pre[1000000 + 5];
  vector<int> p[1000000 + 5];
  void add(int x, int id) {
    pre[id] = id;
    while (x != 1) p[las[x]].push_back(id), x /= las[x];
  }
  void del(int x, int id) {
    pre[id] = id;
    while (x != 1) {
      if (!p[las[x]].size()) {
        pre[id] = 0;
        return;
      }
      pre[id] = min(pre[id], p[las[x]].back());
      p[las[x]].pop_back();
      x /= las[x];
    }
  }
  vector<pair<int, int> > Q;
  int main() {
    int n, i, j;
    scanf("%d", &n);
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    scanf("%s", b + 1);
    get_prime();
    for (i = 1; i <= n; i++)
      if (b[i] == '*')
        add(a[i], i);
      else
        del(a[i], i);
    long long ans = 0;
    for (i = n; i >= 1; i--) {
      int cnt = 1;
      while (!Q.empty() && Q.back().first >= pre[i])
        cnt += Q.back().second, Q.pop_back();
      Q.push_back(make_pair(pre[i], cnt));
      if (pre[i] == i) ans += cnt;
    }
    printf("%lld\n", ans);
  }
  