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 + 5;
  int a[maxn], prim[maxn], tot, flag[maxn], mi[maxn], l[maxn];
  vector<int> st[maxn];
  void init(int n) {
    mi[1] = 1;
    for (int i = 2; i <= n; ++i) {
      if (!flag[i]) prim[tot++] = i, mi[i] = i;
      for (int j = 0; j < tot && prim[j] * i <= n; ++j) {
        flag[i * prim[j]] = 1;
        if (i % prim[j] == 0) {
          mi[i * prim[j]] = mi[i];
          break;
        }
        mi[i * prim[j]] = prim[j];
      }
    }
  }
  char s[maxn];
  int main() {
    init(maxn - 1);
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
    }
    scanf("%s", s + 1);
    for (int i = 1; i <= n; ++i) {
      if (s[i] == '*') {
        l[i] = i;
        int u = a[i];
        while (u != 1) {
          st[mi[u]].push_back(i);
          u /= mi[u];
        }
      } else {
        l[i] = i;
        int u = a[i];
        while (u != 1) {
          if (st[mi[u]].empty()) {
            l[i] = 0;
          } else {
            l[i] = min(l[i], st[mi[u]].back());
            st[mi[u]].pop_back();
          }
          u /= mi[u];
        }
      }
    }
    vector<pair<int, int>> st2;
    long long ans = 0;
    for (int i = n; i >= 1; --i) {
      int cnt = 1;
      while (!st2.empty() && st2.back().first >= i) {
        cnt += st2.back().second;
        st2.pop_back();
      }
      st2.push_back({l[i], cnt});
      if (l[i] == i) ans += cnt;
    }
    printf("%lld\n", ans);
  }
  