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 maxn = 1e6 + 5;
  int f[maxn];
  int n;
  int a[maxn];
  char b[maxn];
  int g[maxn];
  vector<int> stk[maxn], add[maxn];
  void sieve() {
    for (int i = 2; i < maxn; ++i) {
      if (f[i] == 0) {
        for (int j = i; j < maxn; j += i) f[j] = i;
      }
    }
  }
  struct fenwick {
    int sum[maxn];
    void add(int x) {
      for (; x < maxn; x += x & -x) ++sum[x];
    }
    int get(int x) {
      int res = 0;
      for (; x > 0; x -= x & -x) res += sum[x];
      return res;
    }
    int get(int l, int r) { return get(r) - get(l - 1); }
  } tree;
  void read_input() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
    }
    for (int i = 1; i <= n; ++i) {
      cin >> b[i];
    }
  }
  void solve() {
    for (int i = 1; i <= n; ++i) {
      g[i] = i;
      int x = a[i];
      if (b[i] == '*') {
        g[i] = i;
        while (x > 1) {
          stk[f[x]].push_back(i);
          x /= f[x];
        }
      } else {
        while (x > 1) {
          int p = f[x], lst = i;
          bool flag = false;
          while (f[x] == p) {
            if ((int)stk[p].size()) {
              lst = stk[p].back();
              stk[p].pop_back();
            } else
              flag = true;
            x /= p;
          }
          if (flag)
            g[i] = 0;
          else
            g[i] = min(g[i], lst);
        }
      }
      if (g[i]) add[g[i]].push_back(i);
    }
    long long ans = 0;
    for (int i = n; i > 0; --i) {
      for (auto& p : add[i]) tree.add(p);
      int low = i, high = n;
      while (low <= high) {
        int mid = (low + high) / 2;
        if (tree.get(i, mid) == (mid - i + 1))
          low = mid + 1;
        else
          high = mid - 1;
      }
      ans += high - i + 1;
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    sieve();
    read_input();
    solve();
  }
  