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;
  int n, a[N + 9];
  char s[N + 9];
  void into() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    scanf("%s", s + 1);
  }
  struct factor {
    int x, c;
    factor(int X = 0, int C = 0) : x(X), c(C) {}
  };
  vector<factor> d[N + 9];
  void Get_d() {
    for (int i = 2; i <= N; ++i) {
      if (!d[i].empty()) continue;
      for (int j = i; j <= N; j += i) {
        int t = j, c = 0;
        for (; t % i == 0; t /= i) ++c;
        d[j].emplace_back(i, c);
      }
    }
  }
  int uni[N + 9];
  int Query_uni(int k) { return k == uni[k] ? k : uni[k] = Query_uni(uni[k]); }
  struct state {
    int l, r, v;
    state(int L = 0, int R = 0, int V = 0) : l(L), r(R), v(V) {}
    bool operator<(const state &p) const { return v < p.v; }
  };
  priority_queue<state> q[N + 9];
  int last[N + 9], cnt[N + 9];
  long long ans;
  void Get_ans() {
    for (int i = 0; i <= n; ++i) uni[i] = i;
    for (int i = 1, now = 0; i <= n; ++i) {
      ++now;
      for (auto [x, c] : d[a[i]]) {
        q[x].emplace(last[x] + 1, i, cnt[x]);
        cnt[x] += (s[i] == '*' ? 1 : -1) * c;
        last[x] = i;
        for (; !q[x].empty() && q[x].top().v > cnt[x];) {
          auto [l, r, rub] = q[x].top();
          q[x].pop();
          for (int i = Query_uni(r); i >= l; uni[i] = i - 1, i = Query_uni(i))
            --now;
        }
      }
      if (s[i] == '/' && a[i] > 1 && uni[i] == i) uni[i] = i - 1, --now;
      ans += now;
    }
  }
  void work() {
    Get_d();
    Get_ans();
  }
  void outo() { printf("%lld\n", ans); }
  int main() {
    int T = 1;
    for (; T--;) {
      into();
      work();
      outo();
    }
    return 0;
  }
  