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 double eps = 1e-10, pi = 3.1415926535898;
  const int mod = 1e9 + 7, maxn = 1e6 + 10;
  int T, n, m, k, x, y, a[maxn], b[maxn];
  char st[maxn];
  struct P {
    int x, y;
    P(int X, int Y) : x(X), y(Y) {}
    bool operator<(const P& a) const {
      if (x == a.x) return y < a.y;
      return x < a.x;
    }
  } pp(0, 0);
  std::vector<P> v[maxn], p[maxn];
  std::vector<int> tmp[maxn];
  int main(int argc, char const* argv[]) {
    long long ans = 0;
    for (int i = 2; i < maxn; ++i) {
      if (v[i].empty()) {
        for (int j = i; j < maxn; j += i) {
          k = j;
          x = 0;
          while (k % i == 0) {
            k /= i;
            ++x;
          }
          v[j].emplace_back(i, x);
        }
      }
    }
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      b[i] = n;
    }
    scanf("%s", st + 1);
    for (int i = n; i >= 1; --i) {
      if (st[i] == '*') {
        for (P j : v[a[i]]) {
          p[j.x].emplace_back(i, j.y);
        }
      } else {
        for (P j : v[a[i]]) {
          p[j.x].emplace_back(i, -j.y);
        }
      }
    }
    for (int i = 2; i < maxn; ++i) {
      if (!p[i].empty()) {
        p[i].emplace_back(0, 0);
        int su = 0;
        stack<P> s;
        s.push(P(n + 1, 0));
        x = n + 1;
        for (auto j : p[i]) {
          if (s.top().x != n + 1) {
            tmp[j.x + 1].push_back(y);
            tmp[x + 1].push_back(-y);
          }
          if (!j.x and !j.y) {
            break;
          }
          su += j.y;
          if (j.y < 0) {
            s.push(P(j.x, su));
            y = j.x - 1;
            x = j.x;
            continue;
          }
          pp = s.top();
          while (s.top().x != n + 1 and su >= s.top().y) {
            pp = s.top();
            s.pop();
          }
          if (s.top().x == n + 1 and su >= 0) {
            pp = s.top();
          }
          y = pp.x - 1;
          x = j.x;
          if (s.top().x != pp.x) {
            s.push(pp);
          }
          s.push(P(j.x, su));
        }
      }
    }
    multiset<int> ms;
    ms.insert(n);
    for (int i = 1; i <= n; ++i) {
      for (int j : tmp[i]) {
        if (j < 0 or (j == 0 and i > 1)) {
          auto it = ms.find(-j);
          ms.erase(it);
        } else {
          ms.insert(j);
        }
      }
      b[i] = min(b[i], *ms.begin());
    }
    for (int i = 1; i <= n; ++i) {
      ans += b[i] + 1 - i;
    }
    printf("%lld\n", ans);
    return 0;
  }
  