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;
  const int M = 1000005;
  int read() {
    int x = 0, f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9') {
      if (c == '-') f = -1;
    }
    while (c >= '0' && c <= '9') {
      x = (x << 3) + (x << 1) + (c ^ 48);
      c = getchar();
    }
    return x * f;
  }
  int n, m, a[M], b[M], p[M], s[M], num[M];
  vector<int> v[M];
  char t[M];
  long long ans;
  void add(int x, int i) {
    b[i] = i;
    while (x > 1) {
      v[p[x]].push_back(i);
      x /= p[x];
    }
  }
  void del(int x, int i) {
    b[i] = i;
    while (x > 1) {
      if (v[p[x]].empty()) {
        b[i] = 0;
        return;
      }
      b[i] = min(b[i], v[p[x]].back());
      v[p[x]].pop_back();
      x /= p[x];
    }
  }
  signed main() {
    n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    for (int i = 2; i <= 1e6; i++)
      if (!p[i]) {
        p[i] = i;
        for (int j = i; j <= 1e6; j += i) p[j] = i;
      }
    scanf("%s", t + 1);
    for (int i = 1; i <= n; i++) {
      if (t[i] == '*')
        add(a[i], i);
      else
        del(a[i], i);
    }
    for (int i = n; i >= 1; i--) {
      int cnt = 1;
      while (m > 0 && s[m] >= b[i]) {
        cnt += num[m];
        m--;
      }
      s[++m] = b[i];
      num[m] = cnt;
      if (i == b[i]) ans += cnt;
    }
    printf("%lld\n", ans);
  }
  