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>
  const int N = 1000005;
  int n, t;
  inline void read(int &x) {
    register char ch = 0;
    register bool w = 0;
    for (x = 0; !std::isdigit(ch); w |= ch == '-', ch = getchar())
      ;
    for (; std::isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
      ;
    (w) && (x = -x);
    return;
  }
  bool not_prime[N + 5];
  int prime[N], count, minp[N + 5];
  void init() {
    not_prime[1] = 1;
    for (int i = 2; i <= N; ++i) {
      if (!not_prime[i]) prime[++count] = i, minp[i] = i;
      for (int j = 1; j <= count && i * prime[j] <= N; ++j) {
        register int tmp = i * prime[j];
        not_prime[tmp] = 1;
        minp[tmp] = prime[j];
        if (!(i % prime[j])) break;
      }
    }
  }
  char ch[N];
  int ai[N], nxt[N], top;
  std::vector<int> stk[N];
  void add(int val, int loc) {
    while (val > 1) {
      int now = minp[val];
      val /= minp[val];
      stk[now].push_back(loc);
    }
    nxt[loc] = loc;
  }
  void del(int val, int loc) {
    nxt[loc] = loc;
    while (val > 1) {
      int now = minp[val];
      val /= minp[val];
      if (stk[now].empty()) {
        nxt[loc] = 0;
        break;
      }
      nxt[loc] = std::min(nxt[loc], stk[now].back());
      stk[now].pop_back();
    }
  }
  int main() {
    init();
    read(n);
    for (int i = 1; i <= n; ++i) read(ai[i]);
    scanf("%s", ch + 1);
    for (int i = 1; i <= n; ++i) {
      if (ch[i] == '*')
        add(ai[i], i);
      else
        del(ai[i], i);
    }
    static std::pair<int, int> S[N];
    long long ans = 0;
    for (int i = n; i >= 1; --i) {
      while (top && S[top].first >= i) --top;
      if (nxt[i] < i) {
        while (top && S[top].first >= nxt[i]) --top;
        if (nxt[i] < i) S[++top] = std::make_pair(nxt[i], i);
        continue;
      }
      int loc = 0;
      if (top)
        loc = S[top].second - 1;
      else
        loc = n;
      if (loc >= i) ans += loc - i + 1;
    }
    printf("%lld\n", ans);
    return 0;
  }
  