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;
  const int N = 1e6 + 5;
  int read() {
    int s = 0;
    char c = getchar(), lc = '+';
    while (c < '0' || '9' < c) lc = c, c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
    return lc == '-' ? -s : s;
  }
  void write(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10)
      putchar(x + '0');
    else
      write(x / 10), putchar(x % 10 + '0');
  }
  void print(long long x, char c = '\n') {
    write(x);
    putchar(c);
  }
  void up(int &x, int y) {
    if (x > y) x = y;
  }
  int mn[N], a[N], nxt[N];
  void ysgs(int n) {
    mn[1] = 1;
    for (int i = 2; i <= n; i++)
      if (!mn[i])
        for (int j = i; j <= n; j += i)
          if (!mn[j]) mn[j] = i;
  }
  char op[N];
  vector<pair<int, int>> b[N];
  signed main(signed Goodbye, char *Wangang[]) {
    (void)Goodbye, (void)Wangang;
    ysgs(1e6);
    int n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    scanf("%s", op + 1);
    for (int i = 1; i <= n; i++) {
      nxt[i] = i;
      int now = a[i];
      while (now > 1)
        b[mn[now]].push_back(make_pair(op[i] == '*' ? 1 : -1, i)), now /= mn[now];
    }
    for (int t = 1; t <= 1e6; t++)
      if (!b[t].empty()) {
        stack<int> st;
        for (auto i : b[t])
          if (i.first > 0)
            st.push(i.second);
          else if (!st.empty())
            up(nxt[i.second], st.top()), st.pop();
          else
            nxt[i.second] = 0;
      }
    static int st[N], top = 0;
    st[0] = n + 1;
    long long ans = 0;
    for (int i = n; i >= 1; i--) {
      st[++top] = i;
      while (top && nxt[st[top]] >= i) top--;
      ans += st[top] - i;
    }
    print(ans);
    return 0;
  }
  