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 long long inf = (long long)1e18;
  const long long mod = (long long)1e9 + 7;
  const double eps = (double)1e-9;
  const double pi = acos(-1.0);
  const int dx[] = {0, 0, 1, 0, -1};
  const int dy[] = {0, 1, 0, -1, 0};
  const int N = 1000500;
  long long ans;
  vector<pair<int, int> > pr[N];
  int n, a[N], lp[N], pref[N];
  int pw1[N], t[N][20];
  inline void build() {
    for (int i = 2; i <= n; i++) pw1[i] = pw1[i >> 1] + 1;
    for (int j = 0; (1 << j) <= n; j++)
      for (int i = 1; i <= n; i++) t[i][j] = n;
  }
  inline void upd(int l, int r, int val) {
    if (l > r) return;
    int st = pw1[r - l + 1];
    t[l][st] = min(t[l][st], val);
    t[r - (1 << st) + 1][st] = min(t[r - (1 << st) + 1][st], val);
  }
  int main() {
    for (int i = 2; i * i < N; i++) {
      if (!lp[i]) {
        for (int j = i * i; j < N; j += i)
          if (!lp[j]) lp[j] = i;
      }
    }
    for (int i = 2; i < N; i++)
      if (!lp[i]) lp[i] = i;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    getchar();
    for (int i = 1; i <= n; i++) {
      char c = getchar();
      int x = a[i];
      while (x > 1) {
        int p = lp[x], pw = 0;
        while (x % p == 0) {
          pw++;
          x /= p;
        }
        pr[p].push_back({i, (c == '*' ? pw : -pw)});
      }
    }
    build();
    for (int i = 2; i < N; i++) {
      if (pr[i].empty()) continue;
      int m = (int)pr[i].size();
      for (int j = 1; j <= m; j++) pref[j] = pref[j - 1] + pr[i][j - 1].second;
      stack<int> st;
      st.push(m);
      for (int j = m; j >= 1; j--) {
        while (!st.empty() && pref[j - 1] <= pref[st.top()]) st.pop();
        int lim = n;
        if (!st.empty()) {
          int cur = st.top() - 1;
          lim = pr[i][cur].first - 1;
        }
        int beg = (j >= 2 ? pr[i][j - 2].first + 1 : 1);
        upd(beg, pr[i][j - 1].first, lim);
        st.push(j - 1);
      }
    }
    int mx = 0;
    while ((1 << mx) <= n) mx++;
    mx--;
    for (int j = mx; j > 0; j--) {
      for (int l = 1; l <= n - (1 << j) + 1; l++) {
        t[l][j - 1] = min(t[l][j - 1], t[l][j]);
        t[l + (1 << (j - 1))][j - 1] = min(t[l + (1 << (j - 1))][j - 1], t[l][j]);
      }
    }
    for (int i = 1; i <= n; i++) ans += t[i][0] - i + 1;
    printf("%lld", ans);
    return 0;
  }
  