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 <class T>
  void rd(T &x) {
    x = 0;
    char ch = getchar();
    long long f = 1;
    for (; !isdigit(ch); ch = getchar())
      if (ch == '-') f = -1;
    for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + ch - '0';
    x *= f;
  }
  template <class T>
  void cmin(T &x, T y) {
    x = min(x, y);
  }
  template <class T>
  void cmax(T &x, T y) {
    x = max(x, y);
  }
  const long long mod = 1e9 + 7;
  template <class T>
  T add(T x, T y) {
    return (x + y) % mod;
  }
  template <class T>
  T sub(T x, T y) {
    return (x - y + mod) % mod;
  }
  template <class T>
  T mul(T x, T y) {
    return 1ll * x * y % mod;
  }
  template <class T>
  void Add(T &x, T y) {
    x = (x + y) % mod;
  }
  template <class T>
  void Sub(T &x, T y) {
    x = (x - y + mod) % mod;
  }
  template <class T>
  void Mul(T &x, T y) {
    x = 1ll * x * y % mod;
  }
  template <class T>
  T qpow(T a, T b = mod - 2) {
    T ret = 1;
    for (; b; b >>= 1, Mul(a, a))
      if (b & 1) Mul(ret, a);
    return ret;
  }
  const long long N = 1e6 + 10, V = 1e6;
  long long n, a[N], L[N], top;
  pair<long long, long long> stk[N];
  vector<long long> vec[V + 5];
  char s[N];
  long long pri[V + 5], mark[V + 5], cnt, mn[V + 5];
  void sieve(long long n) {
    for (long long i = 2; i <= n; i++) {
      if (!mark[i]) pri[++cnt] = i, mn[i] = i;
      for (long long j = 1; j <= cnt && i * pri[j] <= n; j++) {
        mark[i * pri[j]] = 1, mn[i * pri[j]] = pri[j];
        if (i % pri[j] == 0) break;
      }
    }
  }
  void ins(long long x, long long id) {
    L[id] = id;
    while (x > 1) {
      vec[mn[x]].push_back(id), x /= mn[x];
    }
  }
  void del(long long x, long long id) {
    if (x == 1) {
      L[id] = id;
      return;
    }
    L[id] = id;
    while (x > 1) {
      if (vec[mn[x]].empty()) {
        L[id] = 0;
        return;
      }
      cmin(L[id], vec[mn[x]].back());
      vec[mn[x]].pop_back();
      x /= mn[x];
    }
  }
  signed main() {
    sieve(V);
    rd(n);
    for (long long i = 1; i <= n; i++) rd(a[i]);
    scanf("%s", s + 1);
    for (long long i = 1; i <= n; i++) {
      if (s[i] == '*')
        ins(a[i], i);
      else
        del(a[i], i);
    }
    long long pos = 0;
    long long ans = 0;
    for (long long l = n; l >= 1; l--) {
      long long cnt = 1;
      while (top && stk[top].first >= L[l]) {
        if (pos == top) --pos;
        cnt += stk[top].second;
        --top;
      }
      stk[++top] = make_pair(L[l], cnt);
      if (L[l] == l) ans += cnt;
    }
    printf("%lld\n", ans);
    return 0;
  }
  