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 = 1000005, mx = 1000001;
  int n, a[N], pr[N], tr[N << 2], v[N], st[N], tp;
  char b[N];
  vector<int> fc[N], p[N];
  long long ans;
  inline int cal(int v, int x) {
    int c = 0;
    while (v % x == 0) v /= x, c++;
    return c;
  }
  void build(int k, int l, int r) {
    tr[k] = mx;
    if (l == r) return;
    int mid = l + r >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
  }
  void upd(int k, int l, int r, int a, int b, int v) {
    if (a > b) return;
    if (l == a && r == b) {
      tr[k] = min(tr[k], v);
      return;
    }
    int mid = l + r >> 1;
    if (b <= mid)
      upd(k << 1, l, mid, a, b, v);
    else if (a > mid)
      upd(k << 1 | 1, mid + 1, r, a, b, v);
    else
      upd(k << 1, l, mid, a, mid, v), upd(k << 1 | 1, mid + 1, r, mid + 1, b, v);
  }
  int ask(int k, int l, int r, int x) {
    int v = tr[k];
    if (l == r) return v;
    int mid = l + r >> 1;
    if (x <= mid)
      v = min(v, ask(k << 1, l, mid, x));
    else
      v = min(v, ask(k << 1 | 1, mid + 1, r, x));
    return v;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    scanf("%s", b + 1);
    for (int i = 2; i <= mx; i++)
      for (int j = i + i; j <= mx; j += i) pr[j] = 1;
    for (int i = 2; i <= mx; i++)
      if (!pr[i])
        for (int j = i; j <= mx; j += i) fc[j].push_back(i);
    for (int i = 1; i <= n; i++)
      for (auto j : fc[a[i]]) p[j].push_back(i);
    build(1, 1, n);
    for (int i = 2; i <= mx; i++) {
      if (pr[i]) continue;
      int m = p[i].size();
      if (!m) continue;
      for (int j = 0; j < m; j++)
        v[j] = (j ? v[j - 1] : 0) +
               cal(a[p[i][j]], i) * (b[p[i][j]] == '*' ? 1 : -1);
      tp = 0;
      st[++tp] = m - 1;
      for (int j = m - 1; j >= 0; j--) {
        while (tp && v[st[tp]] >= (j ? v[j - 1] : 0)) tp--;
        if (tp) upd(1, 1, n, j ? p[i][j - 1] + 1 : 1, p[i][j], p[i][st[tp]]);
        st[++tp] = j - 1;
      }
    }
    for (int i = 1; i <= n; i++) ans += min(n + 1, ask(1, 1, n, i)) - i;
    printf("%lld\n", ans);
    return 0;
  }
  