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;
  inline long long read() {
    long long x = 0;
    bool flag = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') flag = 0;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + ch - '0';
      ch = getchar();
    }
    return flag ? x : ~(x - 1);
  }
  inline void write(long long x) {
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
  }
  const long long N = 1e6 + 10;
  long long n, ans, a[N];
  long long prime[N], mp[N], cnt;
  long long stk[N], top, pre[N], w[N];
  bool is_prime[N];
  vector<long long> vec[N];
  inline void sieve(long long x) {
    for (long long i = 2; i <= x; i++) {
      if (!is_prime[i]) prime[++cnt] = mp[i] = i;
      for (long long j = 1; prime[j] <= x / i; j++) {
        is_prime[i * prime[j]] = true;
        mp[i * prime[j]] = prime[j];
        if (i % prime[j] == 0) break;
      }
    }
  }
  signed main() {
    sieve(1000000);
    n = read();
    for (long long i = 1; i <= n; i++) a[i] = read();
    for (long long i = 1; i <= n; i++) {
      char ch = getchar();
      while (ch != '*' && ch != '/') ch = getchar();
      if (ch == '*') {
        while (a[i] > 1) {
          long long p = mp[a[i]];
          vec[p].push_back(i);
          a[i] /= p;
        }
        pre[i] = i;
      } else {
        pre[i] = i;
        while (a[i] > 1) {
          long long p = mp[a[i]];
          if (vec[p].empty())
            pre[i] = -1;
          else
            pre[i] = min(pre[i], vec[p].back()), vec[p].pop_back();
          a[i] /= p;
        }
      }
    }
    for (long long i = n; i >= 1; i--) {
      long long nw = 1;
      while (top && pre[i] <= stk[top]) nw += w[top], top--;
      stk[++top] = pre[i], w[top] = nw;
      if (i == pre[i]) ans += nw;
    }
    write(ans), putchar('\n');
    return 0;
  }
  