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 = 1e6 + 10;
  int n, a[N];
  char s[N];
  bool is[N];
  vector<pair<int, int> > f[N];
  vector<pair<int, int> > id[N];
  long long ans;
  pair<int, int> sta[N];
  int top;
  vector<int> in[N], out[N];
  void work(int x) {
    if (((int)id[x].size()) == 0) return;
    for (auto &y : id[x])
      if (s[y.first] == '/') y.second *= -1;
    for (int i = 1; i < ((int)id[x].size()); i++)
      id[x][i].second += id[x][i - 1].second;
    top = 0;
    for (int i = ((int)id[x].size()) - 1; i >= 0; i--) {
      int now = (i == 0 ? 0 : id[x][i - 1].second);
      while (top && sta[top].second >= id[x][i].second) top--;
      sta[++top] = id[x][i];
      int l = 1, r = top;
      while (l != r) {
        int mid = (l + r) / 2 + 1;
        if (sta[mid].second < now)
          l = mid;
        else
          r = mid - 1;
      }
      if (sta[l].second < now) {
        in[id[x][i].first].push_back(sta[l].first - 1);
        if (i) out[id[x][i - 1].first].push_back(sta[l].first - 1);
      }
    }
  }
  priority_queue<int> pq;
  int tot[N];
  bool vis[N];
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    for (int i = 2; i < N; i++) {
      if (is[i]) continue;
      for (int j = i; j < N; j += i) {
        int u = j, cnt = 0;
        is[j] = 1;
        while (u % i == 0) {
          cnt++;
          u /= i;
        }
        f[j].emplace_back(i, cnt);
      }
    }
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      for (auto x : f[a[i]]) id[x.first].emplace_back(i, x.second);
    }
    cin >> s + 1;
    for (int i = 0; i < N; i++) work(i);
    pq.push(-n);
    tot[n] = 1;
    vis[n] = 1;
    for (int i = n; i >= 1; i--) {
      for (auto x : in[i]) {
        if (!vis[x]) {
          pq.push(-x);
          vis[x] = 1;
        }
        tot[x]++;
      }
      for (auto x : out[i]) tot[x]--;
      while (!pq.empty()) {
        int u = -pq.top();
        if (tot[u] == 0) {
          pq.pop();
          vis[u] = 0;
        } else {
          ans += u - i + 1;
          break;
        }
      }
    }
    cout << ans << endl;
    return 0;
  }
  