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>
  const int MAXN = 1000005;
  using namespace std;
  template <class T>
  using min_heap = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  using max_heap = priority_queue<T, vector<T>, less<T>>;
  int n, A[MAXN], jmp[MAXN];
  int mpf[MAXN];
  vector<pair<int, int>> occ[MAXN];
  void init() {
    for (int i = 1; i < MAXN; i++) mpf[i] = i;
    for (int i = 2; i < MAXN; i++) {
      if (mpf[i] == i) {
        for (int j = i; j < MAXN; j += i) {
          mpf[j] = i;
        }
      }
    }
  }
  void solve() {
    init();
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> A[i];
    }
    for (int i = 0; i < n; i++) {
      char ch;
      cin >> ch;
      int mul = (ch == '*' ? 1 : -1);
      int temp = A[i];
      while (temp != 1) {
        int x = mpf[temp], cnt = 0;
        while (temp % x == 0) {
          temp /= x;
          cnt++;
        }
        occ[x].push_back(pair<int, int>(i, cnt * mul));
      }
    }
    for (int i = 0; i < n; i++) {
      jmp[i] = i;
    }
    for (int p = 2; p < MAXN; p++) {
      if (occ[p].empty()) continue;
      int m = occ[p].size();
      vector<int> sum(m + 1), stk(1);
      for (int i = 0; i < m; i++) {
        sum[i + 1] = sum[i] + occ[p][i].second;
        while (!stk.empty() && sum[stk.back()] > sum[i + 1]) {
          stk.pop_back();
        }
        if (stk.empty()) {
          jmp[occ[p][i].first] = -1;
        } else {
          jmp[occ[p][i].first] =
              min(jmp[occ[p][i].first], occ[p][stk.back()].first);
        }
        stk.push_back(i + 1);
      }
    }
    long long ans = 0;
    jmp[n] = -1e9;
    vector<int> stk(1, n);
    for (int i = n - 1; i >= 0; i--) {
      while (jmp[i] <= jmp[stk.back()]) {
        stk.pop_back();
      }
      if (jmp[i] == i) {
        ans += stk.back() - i;
      }
      stk.push_back(i);
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int tc = 1;
    for (int i = 1; i <= tc; i++) {
      solve();
    }
    return 0;
  }
  