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 maxn = 1e6 + 7;
  long long a[maxn], n;
  char b[maxn];
  long long st[maxn], top, ans, L[maxn];
  int vis[maxn], x, p[maxn], minn[maxn];
  vector<int> v[maxn];
  void oula() {
    for (int i = 2; i < maxn; i++) {
      if (vis[i] == 0) p[++x] = i, minn[i] = i;
      for (int j = 1; j <= x && i * p[j] < maxn; j++) {
        vis[i * p[j]] = 1;
        minn[i * p[j]] = p[j];
        if (i % p[j] == 0) break;
      }
    }
  }
  int main() {
    oula();
    scanf("%lld", &n);
    for (int i = (1); i <= (n); ++i) scanf("%lld", &a[i]);
    scanf("%s", b + 1);
    for (int i = 1; i <= n; i++) {
      long long temp = a[i];
      if (b[i] == '*') {
        L[i] = i;
        while (temp > 1) {
          long long t = minn[temp];
          v[t].push_back(i);
          temp /= t;
        }
        if (temp > 1) v[temp].push_back(i);
      } else {
        int flag = 1, mi = i;
        while (temp > 1) {
          long long t = minn[temp];
          if (v[t].size())
            mi = min(mi, v[t].back()), v[t].pop_back();
          else {
            flag = 0;
          }
          temp /= t;
        }
        if (temp > 1 && v[temp].size())
          mi = min(mi, v[temp].back()), v[temp].pop_back();
        else if (temp > 1 && v[temp].size() == 0)
          flag = 0;
        if (!flag)
          L[i] = 0;
        else
          L[i] = mi;
      }
    }
    st[0] = n + 1;
    top = 0;
    for (int i = n; i >= 1; i--) {
      while (top && L[i] <= L[st[top]]) top--;
      if (i == L[i]) ans += (st[top] - i);
      st[++top] = i;
    }
    printf("%lld", ans);
    return 0;
  }
  