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 = 1 << 21, M = 4e7;
  int n, a[N], pr[N];
  bool Prime[N];
  int p[N >> 3], tot;
  void xxs(int x) {
    for (int i = (2); i <= (x); ++i) {
      if (!Prime[i]) p[++tot] = i, pr[i] = i;
      for (int j = 1; j <= tot && p[j] * i <= x; j++) {
        Prime[p[j] * i] = 1, pr[p[j] * i] = p[j];
        if (i % p[j] == 0) break;
      }
    }
  }
  int ns[N], tg[N];
  void Add(int x, int L, int R, int l, int r, int W) {
    if (l <= L && R <= r) return tg[x] = min(tg[x], W), void();
    int mid = (L + R) >> 1;
    if (l <= mid) Add(x << 1, L, mid, l, r, W);
    if (r > mid) Add(x << 1 | 1, mid + 1, R, l, r, W);
  }
  void Push(int x, int L, int R) {
    if (L == R) return ns[L] = tg[x], void();
    int mid = (L + R) >> 1;
    tg[x << 1] = min(tg[x << 1], tg[x]);
    tg[x << 1 | 1] = min(tg[x << 1 | 1], tg[x]);
    Push(x << 1, L, mid);
    Push(x << 1 | 1, mid + 1, R);
  }
  vector<int> pos[N], c[N];
  char s[N];
  int Pr[M];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    memset(tg, 0x3f, sizeof(tg));
    cin >> n;
    for (int i = (1); i <= (n); ++i) cin >> a[i];
    cin >> (s + 1);
    xxs(1000000);
    for (int i = (1); i <= (n); ++i) {
      int t = a[i];
      while (t > 1) {
        int r = pr[t], cnt = 0;
        while (t % r == 0) t /= r, cnt += 1;
        pos[r].push_back(i);
        if (s[i] == '*')
          c[r].push_back(cnt);
        else
          c[r].push_back(-cnt);
      }
    }
    memset(Pr, 0x3f, sizeof(Pr));
    for (int x = (1); x <= (1000000); ++x) {
      if (!((int)(pos[x]).size())) continue;
      int pn = ((int)(pos[x]).size());
      int sum = 2e7;
      for (int i = (pn - 1); i >= (0); --i) {
        int vr = pn - 1;
        Pr[sum] = min(Pr[sum], i - 1);
        sum += c[x][i];
        for (int t = (sum + 1); t <= (sum + 20); ++t) vr = min(vr, Pr[t]);
        if (vr < pn - 1)
          Add(1, 1, n, (i == 0 ? 1 : pos[x][i - 1] + 1), pos[x][i],
              pos[x][vr + 1] - 1);
      }
      sum = 2e7;
      for (int i = (pn - 1); i >= (0); --i) Pr[sum] = 1e9, sum += c[x][i];
    }
    Push(1, 1, n);
    long long ret = 0;
    for (int i = (1); i <= (n); ++i) ret += min(ns[i], n) - i + 1;
    cout << ret << '\n';
    return 0;
  }
  