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 + 100;
  int n, a[N], p[N], v[N], w, tr[N], st[N], t;
  int s[N], pos[N], nxt[N];
  long long ans;
  char ch[N];
  vector<pair<int, int> > wh[N];
  inline int read() {
    int f = 1, x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9') {
      x = x * 10 + s - '0';
      s = getchar();
    }
    return x * f;
  }
  namespace seg {
  int tag[N * 4];
  void build(int x, int l, int r) {
    tag[x] = n + 1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(x + x, l, mid);
    build(x + x + 1, mid + 1, r);
  }
  void change(int x, int l, int r, int ll, int rr, int v) {
    if (ll <= l && rr >= r) {
      tag[x] = min(tag[x], v);
      return;
    }
    int mid = (l + r) >> 1;
    if (ll <= mid) change(x + x, l, mid, ll, rr, v);
    if (rr > mid) change(x + x + 1, mid + 1, r, ll, rr, v);
  }
  void get(int x, int l, int r, int v) {
    v = min(v, tag[x]);
    if (l == r) {
      ans += v - l;
      return;
    }
    int mid = (l + r) >> 1;
    get(x + x, l, mid, v);
    get(x + x + 1, mid + 1, r, v);
  }
  }  // namespace seg
  void prework() {
    n = 1000000;
    for (int i = 2; i <= n; i++) {
      if (!v[i]) v[i] = p[++w] = i;
      for (int j = 1; j <= w; j++) {
        if (p[j] > n / i || p[j] > v[i]) break;
        v[i * p[j]] = p[j];
      }
    }
    for (int i = 1; i <= w; i++) tr[p[i]] = i;
  }
  signed main() {
    prework();
    n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    scanf("%s", ch + 1);
    for (int i = 1; i <= n; i++) {
      int x = a[i], last = -1, cnt = 0, op = (ch[i] == '/') ? -1 : 1;
      while (x > 1) {
        int now = v[x];
        if (now != last) {
          if (last != -1) wh[tr[last]].emplace_back(i, cnt * op);
          last = now;
          cnt = 0;
        }
        cnt++;
        x /= now;
      }
      if (last != -1) wh[tr[last]].emplace_back(i, cnt * op);
    }
    seg::build(1, 1, n);
    for (int i = 1; i <= w; i++)
      if (!wh[i].empty()) {
        t = 0;
        int m = (int)wh[i].size();
        for (int j = 1; j <= m; j++)
          s[j] = s[j - 1] + wh[i][j - 1].second, pos[j] = wh[i][j - 1].first;
        pos[m + 1] = n + 1;
        for (int j = 0; j <= m; j++) {
          while (t && s[j] < s[st[t]]) nxt[st[t--]] = j;
          st[++t] = j;
        }
        while (t) nxt[st[t--]] = m + 1;
        for (int j = 1; j <= m; j++) {
          if (ch[pos[j]] == '/')
            seg::change(1, 1, n, pos[j - 1] + 1, pos[j], pos[j]);
          else
            seg::change(1, 1, n, pos[j - 1] + 1, pos[j], pos[nxt[j - 1]]);
        }
      }
    seg::get(1, 1, n, 0x3f3f3f3f);
    printf("%lld\n", ans);
  }
  