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 N = 1000000;
  bool primeflag[N + 9];
  const int CP = 78498;
  int Prime[CP], tot, fac[N + 9];
  void prework() {
    for (int i = 2; i <= N; i++) {
      if (!primeflag[i]) Prime[tot] = i, fac[i] = tot, tot++;
      for (int j = 0; j < tot; j++) {
        if (i * Prime[j] > N) break;
        primeflag[i * Prime[j]] = 1, fac[i * Prime[j]] = j;
        if (i % Prime[j] == 0) break;
      }
    }
  }
  stack<int> stk[CP];
  int n, a[N + 9];
  char op[N + 9];
  int pre[N + 9];
  long long Ans;
  struct segment {
    int l, r, val;
  };
  stack<segment> g;
  int main() {
    prework();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", a + i);
    scanf("%s", op + 1);
    for (int i = 1; i <= n; i++) {
      pre[i] = i;
      if (op[i] == '*')
        while (a[i] != 1) stk[fac[a[i]]].push(i), a[i] /= Prime[fac[a[i]]];
      else if (op[i] == '/')
        while (a[i] != 1) {
          if (stk[fac[a[i]]].empty())
            pre[i] = 0;
          else
            pre[i] = min(pre[i], stk[fac[a[i]]].top()), stk[fac[a[i]]].pop();
          a[i] /= Prime[fac[a[i]]];
        }
    }
    for (int i = n; i >= 1; i--) {
      segment now = (segment){i, i, pre[i]};
      while (!g.empty()) {
        segment nxt = g.top();
        if (nxt.val >= pre[i])
          now.r = nxt.r, g.pop();
        else
          break;
      }
      g.push(now);
      if (now.val == i) Ans += now.r - now.l + 1;
    }
    printf("%lld\n", Ans);
    return 0;
  }
  