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 double EPS = 1e-9;
  const int INFMEM = 63;
  const int INF = 1061109567;
  const long long LINF = 4557430888798830399LL;
  const double DINF = numeric_limits<double>::infinity();
  const long long MOD = 1000000007;
  const int dx[8] = {0, 0, 1, -1, 1, -1, 1, -1};
  const int dy[8] = {1, -1, 0, 0, 1, -1, -1, 1};
  const char dch[4] = {'R', 'L', 'D', 'U'};
  const double PI = 3.141592653589793;
  inline void open(string a) {
    freopen((a + ".in").c_str(), "r", stdin);
    freopen((a + ".out").c_str(), "w", stdout);
  }
  inline void fasterios() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  const int MAXN = 1e6;
  int prime[MAXN + 5];
  int isi[MAXN + 5];
  int leftMost[MAXN + 5];
  vector<int> bracket[MAXN + 5];
  int main() {
    fasterios();
    prime[0] = 1;
    prime[1] = 1;
    for (int i = 2; i <= MAXN; i++) {
      if (prime[i]) continue;
      prime[i] = i;
      if (i > 1000) continue;
      for (int j = i * i; j <= MAXN; j += i) prime[j] = i;
    }
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> isi[i];
    }
    for (int i = 1; i <= n; i++) {
      char ch;
      cin >> ch;
      int cur = isi[i];
      leftMost[i] = i;
      if (cur == 1) continue;
      if (ch == '*') {
        leftMost[i] = i + 1;
        while (cur > 1) {
          bracket[prime[cur]].push_back(i);
          cur /= prime[cur];
        }
      } else {
        while (cur > 1) {
          if ((int)(bracket[prime[cur]]).size()) {
            leftMost[i] = min(leftMost[i], bracket[prime[cur]].back());
            bracket[prime[cur]].pop_back();
          } else
            leftMost[i] = 0;
          cur /= prime[cur];
        }
      }
    }
    leftMost[n + 1] = -1;
    stack<int> antri;
    antri.push(n + 1);
    long long ans = 0;
    for (int i = n; i >= 1; i--) {
      while (leftMost[antri.top()] >= i) antri.pop();
      if (leftMost[i] < i) {
        while (leftMost[antri.top()] >= leftMost[i]) antri.pop();
        antri.push(i);
      }
      ans += antri.top() - i;
    }
    cout << ans << '\n';
  }
  