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() {
    char ch = getchar();
    int x = 0, pd = 0;
    while (ch < '0' || ch > '9') pd |= ch == '-', ch = getchar();
    while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
    return pd ? -x : x;
  }
  const int maxn = 1000005;
  const int inf = 0x3f3f3f3f;
  int n, a[maxn];
  char opt[maxn];
  int mx, p[maxn], pl, g[maxn], vis[maxn];
  void pre() {
    for (int i = 2; i <= mx; i++) {
      if (!vis[i]) p[++pl] = g[i] = i;
      for (int j = 1; j <= pl && i * p[j] <= mx; j++) {
        int x = p[j] * i;
        vis[x] = 1, g[x] = p[j];
        if (i % p[j] == 0) break;
      }
    }
  }
  int tim[maxn], T;
  vector<pair<int, int> > stk[maxn];
  int l[maxn];
  namespace SGT {
  int s[maxn << 2];
  void mdf(int o, int l, int r, int x) {
    ++s[o];
    if (l == r) return;
    int mid = l + r >> 1;
    if (x <= mid)
      mdf((o << 1), l, mid, x);
    else
      mdf((o << 1 | 1), mid + 1, r, x);
  }
  void cover(int o, int l, int r, int L, int R) {
    if ((L <= l && r <= R) || !o[s]) {
      o[s] = 0;
      return;
    }
    int mid = l + r >> 1;
    if (L <= mid) cover((o << 1), l, mid, L, R);
    if (R > mid) cover((o << 1 | 1), mid + 1, r, L, R);
    o[s] = (o << 1)[s] + (o << 1 | 1)[s];
  }
  }  // namespace SGT
  using namespace SGT;
  int main() {
    n = read();
    for (int i = 1; i <= n; i++) mx = max(mx, a[i] = read());
    scanf("%s", opt + 1);
    pre();
    for (int i = 1; i <= n; i++)
      if (opt[i] == '*') {
        l[i] = i;
        for (int j = a[i]; j > 1;) {
          int t = g[j], cnt = 0;
          while (g[j] == t) ++cnt, j /= g[j];
          if (tim[t] != T) tim[t] = T, stk[t].clear();
          stk[t].push_back(make_pair(cnt, i));
        }
      } else {
        if (a[i] == 1)
          l[i] = i;
        else
          l[i] = inf;
        for (int j = a[i]; j > 1;) {
          int t = g[j], cnt = 0;
          while (g[j] == t) ++cnt, j /= g[j];
          if (tim[t] != T) tim[t] = T, stk[t].clear();
          while (!stk[t].empty() && cnt > stk[t].back().first)
            cnt -= stk[t].back().first, stk[t].pop_back();
          if (!stk[t].empty()) stk[t].back().first -= cnt, cnt = 0;
          if (cnt) {
            l[i] = 0, ++T;
            break;
          } else
            l[i] = min(l[i], stk[t].back().second);
        }
        if (l[i] == inf) l[i] = 0;
      }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      mdf(1, 1, n, i);
      if (l[i] < i) cover(1, 1, n, l[i] + 1, i);
      ans += s[1];
    }
    printf("%lld\n", ans);
    return 0;
  }
  