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;
  inline char gc() {
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  inline int read() {
    int x = 0;
    char ch = getchar();
    bool positive = 1;
    for (; !isdigit(ch); ch = getchar())
      if (ch == '-') positive = 0;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
    return positive ? x : -x;
  }
  inline void write(long long x) {
    if (x < 0) x = -x, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
  }
  inline void writeln(long long x) {
    write(x);
    puts("");
  }
  struct Tree {
    int minn;
  } st[1000005 * 4];
  int n, m, size, a[1000005], f[1000005], vis[1000005], pri[1000005],
      last[1000005];
  long long ans, g[1000005];
  char s[1000005];
  stack<int> e[100000];
  inline void change(int node, int l, int r, int id, int k) {
    if (l == r) {
      st[node].minn = k;
      return;
    }
    int mid = (l + r) >> 1;
    if (id <= mid) {
      change(node << 1, l, mid, id, k);
    } else {
      change(node << 1 | 1, mid + 1, r, id, k);
    }
    st[node].minn = min(st[node << 1].minn, st[node << 1 | 1].minn);
    return;
  }
  inline int query(int node, int l, int r, int L, int R) {
    if (L >= l && R <= r) return st[node].minn;
    int mid = (L + R) >> 1;
    if (r <= mid) return query(node << 1, l, r, L, mid);
    if (l > mid) return query(node << 1 | 1, l, r, mid + 1, R);
    return min(query(node << 1, l, r, L, mid),
               query(node << 1 | 1, l, r, mid + 1, R));
  }
  int main() {
    n = read();
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      f[i] = i;
    }
    scanf("%s", s + 1);
    for (int i = 1; i <= n; i++)
      if (a[i] == 1) s[i] = '*';
    for (int i = 2; i <= 1000000; i++) {
      if (!vis[i]) {
        pri[++size] = i;
        vis[i] = size;
        last[i] = i;
      }
      for (int j = 1; j <= size && i * pri[j] <= 1000000; j++) {
        int k = i * pri[j];
        vis[k] = 1;
        last[k] = pri[j];
        if (i % pri[j] == 0) break;
      }
    }
    for (int i = 1; i <= n; i++) {
      f[i] = i;
      if (s[i] == '*') {
        while (a[i] != 1) {
          e[vis[last[a[i]]]].push(i);
          a[i] = a[i] / last[a[i]];
        }
      } else {
        while (a[i] != 1) {
          if (e[vis[last[a[i]]]].empty()) {
            f[i] = 0;
          } else {
            f[i] = min(f[i], e[vis[last[a[i]]]].top());
            e[vis[last[a[i]]]].pop();
          }
          a[i] = a[i] / last[a[i]];
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      if (f[i] && f[i] <= i - 1) f[i] = query(1, f[i], i - 1, 1, n);
      change(1, 1, n, i, f[i]);
    }
    g[0] = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] == '*') {
        g[i] = g[i - 1] + 1;
      } else {
        if (!f[i]) continue;
        g[i] = g[f[i] - 1] + 1;
      }
      ans += g[i];
    }
    writeln(ans);
    return 0;
  }
  