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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long N = 1e6 + 5;
  const long long M = 1e9 + 7;
  const long long INF = 0x3f3f3f3f;
  const long long LINF = 0x3f3f3f3f3f3fll;
  const double pi = acos(-1.0);
  long long read() {
    long long a = 0;
    long long d = 1;
    char c;
    while (c = getchar(), c < '0' || c > '9')
      if (c == '-') d = -1;
    a = a * 10 + c - 48;
    while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
    a *= d;
    return a;
  }
  bool num[N];
  long long mi[N];
  long long prime[N];
  long long cnt;
  long long lft[N];
  long long a[N];
  char sen[N];
  vector<long long> q[N];
  void init() {
    num[0] = num[1] = true;
    for (long long i = 2; i < N; i++) {
      if (!num[i]) {
        prime[cnt++] = i;
        mi[i] = i;
      }
      for (long long j = 0; j < cnt && prime[j] * i < N; j++) {
        num[prime[j] * i] = true;
        mi[prime[j] * i] = prime[j];
        if (i % prime[j] == 0) break;
      }
    }
  }
  void add(long long i) {
    lft[i] = i;
    long long x = a[i];
    while (x != 1) {
      long long y = mi[x];
      q[y].push_back(i);
      x /= y;
    }
  }
  void del(long long i) {
    lft[i] = i;
    long long x = a[i];
    while (x != 1) {
      long long y = mi[x];
      if (q[y].empty())
        lft[i] = 0;
      else {
        lft[i] = min(lft[i], q[y].back());
        q[y].pop_back();
      }
      x /= y;
    }
  }
  signed main() {
    init();
    long long n = read();
    for (long long i = 1; i <= n; i++) a[i] = read();
    scanf("%s", sen + 1);
    for (long long i = 1; i <= n; i++) {
      if (sen[i] == '*')
        add(i);
      else
        del(i);
    }
    long long ans = 0;
    stack<long long> stk;
    stk.push(n + 1);
    lft[n + 1] = 0;
    for (long long i = n; i >= 1; i--) {
      while (!stk.empty() && lft[stk.top()] >= lft[i]) stk.pop();
      if (lft[i] == i) ans += stk.top() - i;
      stk.push(i);
    }
    cout << ans;
    return 0;
  }
  