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;
  struct qt {
    int val, id;
    qt(int val = 0, int id = 0) : val(val), id(id) {}
    bool operator<(const qt q) const { return val > q.val; }
  };
  int n, a[1000005], p[1000005], g[1000005], prime[1000005], pt,
      pn = 1e6, mn[1000005], mx[1000005];
  int c[1000005], nans[1000005];
  char s[1000005];
  vector<int> val[1000005];
  priority_queue<qt> pq;
  void modify(int x, int v, int i) {
    val[x][c[x]] = i;
    c[x] -= v;
    nans[x] = val[x][c[x] - 1];
    pq.push(qt(nans[x], x));
  }
  int getans() {
    while (!pq.empty() && nans[pq.top().id] != pq.top().val) pq.pop();
    return pq.top().val;
  }
  int main() {
    p[1] = 1;
    for (int i = 2; i <= pn; i++) {
      if (!p[i]) prime[++pt] = i, g[i] = i;
      for (int j = 1; j <= pt && i * prime[j] <= pn; j++) {
        p[i * prime[j]] = 1;
        g[i * prime[j]] = prime[j];
        if (i % prime[j] == 0) break;
      }
    }
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    scanf("%s", s + 1);
    for (int i = n; i > 0; i--) {
      int u = a[i];
      while (u != 1) {
        int v = g[u];
        if (s[i] == '*')
          c[v]--;
        else
          c[v]++;
        mn[v] = min(mn[v], c[v]);
        mx[v] = max(mx[v], c[v]);
        u /= v;
      }
    }
    for (int i = 1; i <= pn; i++)
      if (!p[i]) {
        val[i].resize(mx[i] - mn[i] + 3);
        for (int j = 0; j <= mx[i] - mn[i] + 2; j++) val[i][j] = n + 1;
        c[i] = -mn[i] + 1;
        nans[i] = n + 1;
        pq.push(qt(nans[i], i));
      }
    long long ans = 0;
    for (int i = n; i > 0; i--) {
      int u = a[i];
      while (u != 1) {
        int v = g[u];
        if (s[i] == '*')
          modify(v, 1, i);
        else
          modify(v, -1, i);
        u /= v;
      }
      ans += getans() - i;
    }
    printf("%lld\n", ans);
    return 0;
  }
  