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_MAX = 1000000;
  const int A_MAX = 1000000;
  int pdiv[A_MAX + 2];
  int n;
  int a[N_MAX + 2];
  int sgn[N_MAX + 2];
  vector<pair<int, short int>> muls[A_MAX + 2];
  int SGT[N_MAX * 4 + 2];
  void build(int node, int l, int r) {
    SGT[node] = n;
    if (l == r) return;
    int mid = (l + r) / 2;
    int lSon = node * 2, rSon = node * 2 + 1;
    build(lSon, l, mid);
    build(rSon, mid + 1, r);
  }
  void build() { build(1, 1, n); }
  void update(int node, int l, int r, int ul, int ur, int uval) {
    if (ul <= l && r <= ur) {
      SGT[node] = min(SGT[node], uval);
      return;
    }
    int mid = (l + r) / 2;
    int lSon = node * 2, rSon = node * 2 + 1;
    if (ul <= mid) update(lSon, l, mid, ul, ur, uval);
    if (mid + 1 <= ur) update(rSon, mid + 1, r, ul, ur, uval);
  }
  void update(int ul, int ur, int uval) { update(1, 1, n, ul, ur, uval); }
  int query(int node, int l, int r, int qpos) {
    if (l == r) return SGT[node];
    int mid = (l + r) / 2;
    int lSon = node * 2, rSon = node * 2 + 1;
    if (qpos <= mid)
      return min(SGT[node], query(lSon, l, mid, qpos));
    else
      return min(SGT[node], query(rSon, mid + 1, r, qpos));
  }
  int query(int qpos) { return query(1, 1, n, qpos); }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    for (int p = 2; p <= A_MAX; p++)
      if (pdiv[p] == 0)
        for (int m = p; m <= A_MAX; m += p) pdiv[m] = p;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    string str;
    cin >> str;
    for (int i = 1; i <= n; i++) sgn[i] = (str[i - 1] == '*' ? +1 : -1);
    for (int i = 1; i <= n; i++) {
      int aux = a[i];
      while (aux > 1) {
        int p = pdiv[aux];
        int e = 0;
        do {
          e++;
          aux /= p;
        } while (aux % p == 0);
        muls[p].push_back(make_pair(i, e * sgn[i]));
      }
    }
    build();
    for (int p = 1; p <= A_MAX; p++)
      if (muls[p].empty() == false) {
        vector<int> pref;
        for (int k = 0; k < (int)muls[p].size(); k++) {
          pref.push_back((pref.empty() == true ? 0 : pref.back()));
          pref.back() += muls[p][k].second;
        }
        vector<int> st;
        for (int k = (int)muls[p].size() - 1; k >= 0; k--) {
          while (st.empty() == false && pref[k] <= pref[st.back()]) st.pop_back();
          int l = muls[p][k].first + 1;
          int r = (st.empty() == true ? n + 1 : muls[p][st.back()].first);
          update(l, r, r - 1);
          st.push_back(k);
        }
        while (st.empty() == false && 0 <= pref[st.back()]) st.pop_back();
        int l = 1;
        int r = (st.empty() == true ? n + 1 : muls[p][st.back()].first);
        update(l, r, r - 1);
      }
    long long answer = 0;
    for (int l = 1; l <= n; l++) answer += max(0, query(l) - l + 1);
    cout << answer << "\n";
    return 0;
  }
  