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 std::pair;
  using std::vector;
  const int MAX_N = 1000005, MAX_V = 1000005, MAX_PN = 78499;
  int v[MAX_V], p[MAX_PN], pn = 0;
  int a[MAX_N];
  char opt[MAX_N];
  int req[MAX_N];
  int n;
  vector<int> vec[MAX_V];
  pair<int, int> stk[MAX_N];
  inline void chkmn(int &a, int b) {
    if (a > b) a = b;
  }
  void eulerSieve(int n = 1000000) {
    for (int i = 2; i <= n; ++i) {
      if (!v[i]) v[i] = p[++pn] = i;
      for (int j = 1; 1ll * i * p[j] <= n; ++j) {
        v[i * p[j]] = p[j];
        if (i % p[j] == 0) break;
      }
    }
  }
  void ins(int pos) {
    req[pos] = pos;
    for (int x = a[pos]; x > 1; x /= v[x]) vec[v[x]].push_back(pos);
  }
  void del(int pos) {
    req[pos] = pos;
    for (int x = a[pos]; x > 1; x /= v[x]) {
      if (vec[v[x]].empty()) {
        req[pos] = 0;
        return;
      }
      chkmn(req[pos], vec[v[x]].back());
      vec[v[x]].pop_back();
    }
  }
  int main() {
    eulerSieve();
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", a + i);
    for (int i = 1; i <= n; ++i) scanf(" %c", opt + i);
    for (int i = 1; i <= n; ++i)
      if (opt[i] == '*')
        ins(i);
      else
        del(i);
    long long ans = 0;
    for (int i = n, top = 0; i; --i) {
      int cnt = 1;
      while (top and req[i] <= stk[top].first) cnt += stk[top--].second;
      stk[++top] = pair<int, int>(req[i], cnt);
      if (req[i] == i) ans += cnt;
    }
    printf("%lld\n", ans);
    return 0;
  }
  