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;
  template <class T>
  bool ckmin(T &a, const T &b) {
    return a > b ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T &a, const T &b) {
    return a < b ? a = b, 1 : 0;
  }
  namespace debug {
  void __print(int x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cerr << '{';
    for (auto z : x) cerr << (f++ ? "," : ""), __print(z);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  }  // namespace debug
  using namespace debug;
  const char nl = '\n';
  mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
  const int mxN = 1e6 + 12;
  int lp[mxN];
  void precalc() {
    vector<int> primes;
    for (int i = 2; i < mxN; ++i) {
      if (lp[i] == 0) {
        lp[i] = i;
        primes.push_back(i);
      }
      for (int j = 0; j < (int)(primes).size() && primes[j] <= lp[i] &&
                      primes[j] * 1LL * i < mxN;
           ++j) {
        lp[primes[j] * i] = primes[j];
      }
    }
  }
  vector<int> gde[mxN];
  const int LOG = 24;
  int sparse[mxN][LOG];
  void solve() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto &z : v) cin >> z;
    string s;
    cin >> s;
    vector<int> granica(n);
    for (int i = 0; i < n; ++i) {
      int x = v[i];
      if (s[i] == '*') {
        granica[i] = i;
        while (x > 1) {
          gde[lp[x]].push_back(i);
          x /= lp[x];
        }
      } else {
        granica[i] = i;
        while (x > 1) {
          int tren = lp[x];
          if ((int)(gde[tren]).size() == 0) {
            granica[i] = -1;
            x /= tren;
            continue;
          }
          ckmin(granica[i], gde[tren].back());
          gde[tren].pop_back();
          x /= tren;
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      sparse[i][0] = granica[i];
    }
    for (int iter = 1; iter < LOG; ++iter) {
      for (int i = 0; i + (1 << iter) - 1 < n; ++i) {
        int drugi = (1 << (iter - 1)) + i;
        sparse[i][iter] = min(sparse[i][iter - 1], sparse[drugi][iter - 1]);
      }
    }
    auto query = [&](int l, int r) {
      int duz = r - l + 1;
      int treba = 31 - __builtin_clz(duz);
      int levi = l;
      int desni = r - (1 << treba) + 1;
      return min(sparse[levi][treba], sparse[desni][treba]);
    };
    long long odg = 0;
    for (int i = 0; i < n; ++i) {
      int l = i, r = n - 1;
      int ans = -1;
      while (l <= r) {
        int mid = (l + r) / 2;
        if (query(i, mid) == i) {
          ans = mid;
          l = mid + 1;
        } else {
          r = mid - 1;
        }
      }
      if (ans >= i) {
        odg += (ans - i + 1);
      }
    }
    cout << odg << nl;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    precalc();
    int testCases = 1;
    while (testCases--) solve();
  }
  