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;
  template <typename S, typename T>
  bool minx(S &a, const T &b) {
    return a < b ? 1 : a = b, 0;
  }
  template <typename T>
  void int_in(T &res) {
    int flag = 1;
    char ch;
    if ((ch = getchar()) == '-')
      flag = -1;
    else if (ch >= '0' && ch <= '9')
      res = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - '0';
    res *= flag;
  }
  void int_out(long long a) {
    if (a > 9) int_out(a / 10);
    putchar(a % 10 + '0');
  }
  const int INF = 0x3f3f3f3f;
  const double tol = 1e-8;
  const int N = 1000005;
  int primes[N], minp[N], a[N], left_pos[N];
  int liner_sieve(int n, int primes[], int minp[]) {
    int cnt = 0;
    memset(minp, 0, n * sizeof *minp);
    for (int i = 2; i < n; ++i) {
      if (!minp[i]) {
        minp[i] = primes[cnt++] = i;
      }
      for (int j = 0; primes[j] * i < n; ++j) {
        minp[primes[j] * i] = primes[j];
        if (i % primes[j] == 0) break;
      }
    }
    return cnt;
  }
  vector<int> pos[N];
  void add(int idx, int num) {
    left_pos[idx] = idx;
    for (int p = minp[num]; num != 1; p = minp[num /= p]) pos[p].push_back(idx);
  }
  void del(int idx, int num) {
    left_pos[idx] = idx;
    for (int p = minp[num]; num != 1; p = minp[num /= p]) {
      if (pos[p].size() == 0) {
        left_pos[idx] = 0;
        return;
      }
      minx(left_pos[idx], pos[p].back());
      pos[p].pop_back();
    }
  }
  int main() {
    int n;
    int_in(n);
    liner_sieve(N, primes, minp);
    for (int i = 1; i <= n; ++i) int_in(a[i]);
    char type;
    for (int i = 1; i <= n; ++i) {
      scanf("%c", &type);
      if (type == '*')
        add(i, a[i]);
      else
        del(i, a[i]);
    }
    vector<int> stk;
    left_pos[n + 1] = -1;
    stk.push_back(n + 1);
    long long ans = 0;
    for (int i = n; i >= 1; --i) {
      while (left_pos[stk.back()] >= left_pos[i]) {
        stk.pop_back();
      }
      if (left_pos[i] == i) ans += stk.back() - i;
      stk.push_back(i);
    }
    printf("%lld\n", ans);
    return 0;
  }
  