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;
  int read() {
    int a = 0;
    int f = 0;
    char p = getchar();
    while (!isdigit(p)) {
      f |= p == '-';
      p = getchar();
    }
    while (isdigit(p)) {
      a = (a << 3) + (a << 1) + (p ^ 48);
      p = getchar();
    }
    return f ? -a : a;
  }
  const int INF = 998244353;
  int T;
  int n;
  long long ans;
  int val[1000050];
  char s[1000050];
  int pre[1000050];
  bool vis[1000050];
  int t[1000050], top;
  int sum[1000050];
  int pos[1000050];
  vector<pair<int, int> > q[1000050];
  int bl[4000050];
  void init() {
    for (int i = 1; i <= n * 4; ++i) bl[i] = n + 1;
    for (int i = 2; i <= 1000000; ++i) {
      if (!vis[i]) pre[i] = i, t[++top] = i;
      for (int j = 1; j <= top && i * t[j] <= 1000000; ++j) {
        vis[i * t[j]] = true;
        pre[i * t[j]] = t[j];
      }
    }
  }
  void modify(int u, int l, int r, int L, int R, int V) {
    if (L <= l && r <= R) {
      bl[u] = min(bl[u], V);
      return;
    }
    int mid = (l + r) >> 1;
    if (L <= mid) modify(u << 1, l, mid, L, R, V);
    if (R > mid) modify(u << 1 | 1, mid + 1, r, L, R, V);
  }
  void calc(int u, int l, int r, int v) {
    v = min(v, bl[u]);
    if (l == r) {
      ans += v - l;
      return;
    }
    int mid = (l + r) >> 1;
    calc(u << 1, l, mid, v);
    calc(u << 1 | 1, mid + 1, r, v);
  }
  int main() {
    n = read();
    init();
    for (int i = 1; i <= n; ++i) val[i] = read();
    scanf("%s", s + 1);
    for (int i = 1; i <= n; ++i) {
      int tmp = val[i];
      while (tmp != 1) {
        int now = pre[tmp];
        int cnt = 0;
        while (pre[tmp] == now) tmp /= now, cnt++;
        if (s[i] == '/') cnt = -cnt;
        q[now].push_back({i, cnt});
      }
    }
    for (int k = 2; k <= 1000000; ++k) {
      if (vis[k]) continue;
      int m = q[k].size();
      if (!m) continue;
      for (int i = 1; i <= m; ++i) {
        pos[i] = q[k][i - 1].first;
        sum[i] = sum[i - 1] + q[k][i - 1].second;
      }
      pos[m + 1] = n + 1;
      sum[m + 1] = -INF;
      top = 0;
      t[++top] = m + 1;
      for (int i = m; i >= 1; --i) {
        t[++top] = i;
        while (sum[t[top]] >= sum[i - 1]) top--;
        modify(1, 1, n, pos[i - 1] + 1, pos[i], pos[t[top]]);
      }
    }
    calc(1, 1, n, n + 1);
    printf("%lld", ans);
    return 0;
  }
  