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 ma = 1e6, inf = 1e9;
  int n, nn[1000069], fq[1000069], a[1000069];
  pair<int, int> fz[1000069][7];
  vector<pair<int, int>> sk[1000069];
  bitset<1000069> spc;
  multiset<int> ms;
  int main() {
    int i, j, k, w, c, u, ww;
    long long z = 0;
    char ch;
    for (i = 2; i <= ma; i++) {
      if (!fq[i]) {
        for (j = i; j <= ma; j += i) {
          for (c = 0, k = j; k % i == 0; k /= i, c++)
            ;
          fz[j][fq[j]] = {i, c};
          fq[j]++;
        }
        sk[i].push_back({-1, -inf});
        sk[i].push_back({0, 0});
        nn[i] = 1;
        ms.insert(-1);
      }
    }
    scanf("%d", &n);
    for (i = n; i; i--) {
      scanf("%d", a + i);
    }
    for (i = n; i; i--) {
      scanf(" %c", &ch);
      spc[i] = ch == '/';
    }
    for (i = 1; i <= n; i++) {
      u = spc[i] * 2 - 1;
      for (j = 0; j < fq[a[i]]; j++) {
        k = fz[a[i]][j].first;
        w = fz[a[i]][j].second;
        sk[k][nn[k]].first = i - 1;
        ms.erase(ms.find(sk[k][nn[k] - 1].first));
        ww = sk[k][nn[k]].second + w * u;
        for (; nn[k] && sk[k][nn[k]].second >= ww; sk[k].pop_back(), nn[k]--)
          ;
        ms.insert(sk[k][nn[k]].first);
        sk[k].push_back({i, ww});
        nn[k]++;
      }
      z += i - 1 - *prev(ms.end());
    }
    printf("%lld\n", z);
  }
  