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;
  bool home = 1;
  const int N = (int)1e6 + 7;
  int n, a[N], b[N], lim[4 * N];
  vector<int> primes;
  int lp[N];
  void build(int v, int tl, int tr) {
    lim[v] = n;
    if (tl < tr) {
      int tm = (tl + tr) / 2;
      build(2 * v, tl, tm);
      build(2 * v + 1, tm + 1, tr);
    }
  }
  void force(int v, int tl, int tr, int l, int r, int x) {
    if (tr < l || r < tl) return;
    if (l <= tl && tr <= r) {
      lim[v] = min(lim[v], x);
      return;
    }
    int tm = (tl + tr) / 2;
    force(2 * v, tl, tm, l, r, x);
    force(2 * v + 1, tm + 1, tr, l, r, x);
  }
  void force(int l, int r, int x) { force(1, 1, n, l, r, x); }
  int get(int v, int tl, int tr, int i) {
    if (tr < i || i < tl) return n;
    int sol = lim[v];
    if (tl < tr) {
      int tm = (tl + tr) / 2;
      sol = min(sol, get(2 * v, tl, tm, i));
      sol = min(sol, get(2 * v + 1, tm + 1, tr, i));
    }
    return sol;
  }
  vector<pair<int, int>> cool[N];
  int cur[N], suf[N], first[N];
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    home = 0;
    if (home) {
      freopen("input", "r", stdin);
    }
    for (int i = 2; i < N; 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] * i < N;
           j++) {
        lp[primes[j] * i] = primes[j];
      }
    }
    cin >> n;
    build(1, 1, n);
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      int num = a[i];
      while (num > 1) {
        int p = lp[num], cnt = 0;
        while (num % p == 0) {
          num /= p;
          cnt++;
        }
        cool[p].push_back({i, cnt});
      }
    }
    {
      string s;
      cin >> s;
      for (int i = 1; i <= n; i++) {
        char ch = s[i - 1];
        assert(ch == '*' || ch == '/');
        if (ch == '*')
          b[i] = 1;
        else
          b[i] = 2;
      }
    }
    for (int p = 1; p < N; p++) {
      if (cool[p].empty()) continue;
      assert(lp[p] == p);
      vector<int> pos;
      for (auto &it : cool[p]) {
        pos.push_back(it.first);
        if (b[it.first] == 1)
          cur[it.first] = it.second;
        else
          cur[it.first] = -it.second;
      }
      int y = (int)pos.size();
      vector<pair<int, int>> segs;
      if (pos[0] > 1) {
        segs.push_back({1, pos[0] - 1});
      }
      for (int i = 0; i + 1 < y; i++) {
        segs.push_back({pos[i], pos[i + 1] - 1});
      }
      segs.push_back({pos[y - 1], n});
      {
        int sz = 0;
        for (auto &it : segs) sz += it.second - it.first + 1;
        assert(sz == n);
        assert(segs[0].first == 1);
        assert(segs.back().second == n);
        for (int i = 0; i + 1 < (int)segs.size(); i++) {
          assert(segs[i].second + 1 == segs[i + 1].first);
        }
      }
      y = (int)segs.size();
      suf[y] = 0;
      for (int i = y - 1; i >= 0; i--) {
        suf[i] = suf[i + 1] + cur[segs[i].first];
      }
      vector<int> stk;
      for (int i = y; i >= 0; i--) {
        while (!stk.empty() && suf[i] >= suf[stk.back()]) stk.pop_back();
        if (stk.empty())
          first[i] = -1;
        else
          first[i] = stk.back();
        stk.push_back(i);
      }
      for (int i = y - 1; i >= 0; i--) {
        if (first[i] != -1) {
          int j = first[i];
          assert(suf[i] < suf[j]);
          force(segs[i].first, segs[i].first, segs[j - 1].first - 1);
        }
        if (segs[i].first + 1 <= segs[i].second) {
          if (first[i + 1] != -1) {
            int j = first[i + 1];
            force(segs[i].first + 1, segs[i].second, segs[j - 1].first - 1);
          }
        }
      }
      for (auto &it : cool[p]) {
        cur[it.first] = 0;
      }
    }
    long long sol = 0;
    for (int i = 1; i <= n; i++) {
      int r = get(1, 1, n, i);
      if (r >= i) {
        sol += r - i + 1;
      }
    }
    cout << sol << "\n";
    return 0;
  }
  