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 INF = 100000000;
  const int MAX = 1000000;
  template <typename T>
  struct dual_segment_tree {
    int N;
    vector<T> ST;
    dual_segment_tree(int n, int x) {
      N = 1;
      while (N < n) {
        N *= 2;
      }
      ST = vector<T>(N * 2 - 1, x);
    }
    T operator[](int k) {
      k += N - 1;
      T ans = ST[k];
      while (k > 0) {
        k = (k - 1) / 2;
        ans = min(ans, ST[k]);
      }
      return ans;
    }
    void range_chmin(int L, int R, T x, int i, int l, int r) {
      if (R <= l || r <= L) {
        return;
      } else if (L <= l && r <= R) {
        ST[i] = min(ST[i], x);
        return;
      } else {
        int m = (l + r) / 2;
        range_chmin(L, R, x, i * 2 + 1, l, m);
        range_chmin(L, R, x, i * 2 + 2, m, r);
        return;
      }
    }
    void range_chmin(int L, int R, T x) { range_chmin(L, R, x, 0, 0, N); }
  };
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<char> b(n);
    for (int i = 0; i < n; i++) {
      cin >> b[i];
    }
    vector<int> prime(MAX + 1, -1);
    for (int i = 2; i <= MAX; i++) {
      if (prime[i] == -1) {
        for (int j = i * 2; j <= MAX; j += i) {
          prime[j] = i;
        }
      }
    }
    vector<vector<pair<int, int>>> id(MAX);
    for (int i = 0; i < n; i++) {
      int x = a[i];
      while (prime[x] != -1) {
        int p = prime[x];
        int cnt = 0;
        while (x % p == 0) {
          x /= p;
          cnt++;
        }
        if (b[i] == '*') {
          id[p].push_back(make_pair(cnt, i));
        } else {
          id[p].push_back(make_pair(-cnt, i));
        }
      }
      if (x > 1) {
        if (b[i] == '*') {
          id[x].push_back(make_pair(1, i));
        } else {
          id[x].push_back(make_pair(-1, i));
        }
      }
    }
    dual_segment_tree<int> ST(n, n);
    for (int i = 0; i < MAX; i++) {
      if (!id[i].empty()) {
        int cnt = id[i].size();
        stack<pair<int, int>> st;
        st.push(make_pair(-INF, n + 1));
        st.push(make_pair(0, n));
        for (int j = cnt - 1; j >= 0; j--) {
          int p = id[i][j].second;
          int pp;
          if (j == 0) {
            pp = -1;
          } else {
            pp = id[i][j - 1].second;
          }
          if (id[i][j].first < 0) {
            ST.range_chmin(pp + 1, p + 1, p);
            int nxt = st.top().first - id[i][j].first;
            st.push(make_pair(nxt, p));
          } else {
            int nxt = st.top().first - id[i][j].first;
            while (st.size() >= 2) {
              pair<int, int> P = st.top();
              st.pop();
              if (st.top().first < nxt) {
                st.push(P);
                break;
              }
            }
            int r = st.top().second;
            ST.range_chmin(pp + 1, p + 1, r);
            st.push(make_pair(nxt, p));
          }
        }
      }
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      ans += ST[i] - i;
    }
    cout << ans << endl;
  }
  