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 <typename T>
  void chkMax(T &x, T y) {
    if (y > x) x = y;
  }
  template <typename T>
  void chkMin(T &x, T y) {
    if (y < x) x = y;
  }
  template <typename T>
  void inline read(T &x) {
    int f = 1;
    x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
  }
  const int N = 1e6 + 5;
  int n, a[N], p0[N];
  void inline prework() {
    int n = 1e6;
    for (int i = 2; i <= n; i++) {
      for (int j = i; j <= n; j += i)
        if (!p0[j]) p0[j] = i;
    }
  }
  vector<pair<int, int> > e[N];
  char g[N];
  int s[N], top, sum[N];
  vector<pair<int, int> > q[N];
  vector<pair<int, int> > G;
  void inline work(vector<pair<int, int> > &t) {
    int m = t.size();
    if (!m) return;
    G.clear();
    G.push_back(make_pair(0, 0));
    for (int i = 0; i < m; i++) G.push_back(t[i]);
    for (int i = 1; i <= m; i++) {
      sum[i] = G[i].first + sum[i - 1];
    }
    top = 0;
    for (int i = m; i >= 0; i--) {
      while (top && sum[s[top]] >= sum[i]) top--;
      if (s[top]) {
        q[G[i].second + 1].push_back(make_pair(G[s[top]].second, 1));
        if (i != m)
          q[G[i + 1].second + 1].push_back(make_pair(G[s[top]].second, -1));
      }
      s[++top] = i;
    }
  }
  multiset<int> S;
  int main() {
    prework();
    read(n);
    for (int i = 1; i <= n; i++) read(a[i]);
    scanf("%s", g + 1);
    for (int i = 1; i <= n; i++) {
      int x = a[i];
      int w = g[i] == '*' ? 1 : -1;
      while (x > 1) {
        int v = p0[x], c = 0;
        while (x % v == 0) x /= v, c++;
        e[v].push_back(make_pair(w * c, i));
      }
    }
    long long ans = 0;
    for (int i = 1; i <= 1000000; i++) work(e[i]);
    for (int i = 1; i <= n; i++) {
      for (pair<int, int> u : q[i]) {
        if (u.second == 1)
          S.insert(u.first);
        else
          S.erase(S.find(u.first));
      }
      int lim = n;
      if (S.size()) lim = *S.begin() - 1;
      if (i <= lim) ans += lim - i + 1;
    }
    printf("%lld\n", ans);
    return 0;
  }
  