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, A = 1e6 + 5;
  int n, a[N], nxt[N];
  unordered_map<int, stack<int> > buc;
  bool tag[A];
  int p[A];
  vector<int> pri;
  long long res = 0;
  void init() {
    for (int i = 2; i < A; ++i) {
      if (!tag[i]) pri.push_back(i), p[i] = i;
      for (int j = 0; j < (int)pri.size(); ++j) {
        if (i * pri[j] >= A) break;
        tag[i * pri[j]] = true, p[i * pri[j]] = pri[j];
        if (i % pri[j] == 0) break;
      }
    }
  }
  void add(int i, int x) {
    nxt[i] = i;
    while (x > 1) buc[p[x]].push(i), x /= p[x];
  }
  void del(int i, int x) {
    nxt[i] = i;
    while (x > 1) {
      if (buc[p[x]].empty())
        nxt[i] = 0;
      else
        nxt[i] = min(nxt[i], buc[p[x]].top()), buc[p[x]].pop();
      x /= p[x];
    }
  }
  struct Seg_Tree {
    struct Node {
      int data;
    } tr[N << 2];
    void up(int u) { tr[u].data = min(tr[u << 1].data, tr[u << 1 | 1].data); }
    void modify(int u, int l, int r, int x, int z) {
      if (l == r) return tr[u].data = z, void();
      int mid = (l + r) >> 1;
      if (x <= mid)
        return modify(u << 1, l, mid, x, z), up(u);
      else
        return modify(u << 1 | 1, mid + 1, r, x, z), up(u);
    }
    int query(int u, int l, int r, int x, int y, int z) {
      int mid = (l + r) >> 1, res = n + 1;
      if (x <= l && r <= y) {
        if (tr[u].data >= z) return n + 1;
        if (l == r) return l;
        if (tr[u << 1].data < z)
          return query(u << 1, l, mid, x, y, z);
        else
          return query(u << 1 | 1, mid + 1, r, x, y, z);
      }
      if (x <= mid) res = query(u << 1, l, mid, x, y, z);
      if (res != n + 1) return res;
      if (y > mid) res = query(u << 1 | 1, mid + 1, r, x, y, z);
      return res;
    }
  } t;
  int main() {
    init(), cin >> n;
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= n; ++i) {
      char c = getchar();
      while (c != '*' && c != '/') c = getchar();
      if (c == '*')
        add(i, a[i]);
      else
        del(i, a[i]);
    }
    for (int i = 1; i <= n; ++i) t.modify(1, 1, n, i, nxt[i]);
    for (int i = 1; i <= n; ++i) res += t.query(1, 1, n, i, n, i) - i;
    return printf("%lld\n", res), 0;
  }
  