Problem:
Score : 600 points

Problem Statement

Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.

There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.

Notes

A string a is a substring of another string b if and only if there exists an integer x (0 \leq x \leq |b| - |a|) such that, for any y (1 \leq y \leq |a|), a_y = b_{x+y} holds.

We assume that the concatenation of zero copies of any string is the empty string. From the definition above, the empty string is a substring of any string. Thus, for any two strings s and t, i = 0 satisfies the condition in the problem statement.

Constraints

1 \leq |s| \leq 5 \times 10^5

1 \leq |t| \leq 5 \times 10^5

s and t consist of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print -1.

Sample Input 1

abcabab
ab

Sample Output 1

3

The concatenation of three copies of t, ababab, is a substring of the concatenation of two copies of s, abcabababcabab, so i = 3 satisfies the condition.

On the other hand, the concatenation of four copies of t, abababab, is not a substring of the concatenation of any number of copies of s, so i = 4 does not satisfy the condition.

Similarly, any integer greater than 4 does not satisfy the condition, either. Thus, the number of non-negative integers i satisfying the condition is finite, and the maximum value of such i is 3.

Sample Input 2

aa
aaaaaaa

Sample Output 2

-1

For any non-negative integer i, the concatenation of i copies of t is a substring of the concatenation of 4i copies of s. Thus, there are infinitely many non-negative integers i that satisfy the condition.

Sample Input 3

aba
baaab

Sample Output 3

0

As stated in Notes, i = 0 always satisfies the condition.

Solution in Rust (1.15.1) (judge verdict: Time Limit Exceeded)

use std::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let s: String = input();
    let t: String = input();

    let ns = s.len();
    let nt = t.len();
    let mut u = s.clone();
    while u.len() < ns + nt {
        u.push_str(&s);
    }

    let mut ok = vec![false; ns];
    for i in 0..ns {
        if u[i..i + nt] == t {
            ok[i] = true;
        }
    }

    let mut cycle = false;
    let mut uf = UnionFind::new(ns);
    for i in 0..ns {
        let j = (i + nt) % ns;
        if ok[i] && ok[j] {
            if !uf.union(i, j) {
                cycle = true;
            }
        }
    }

    if cycle {
        println!("-1");
    } else {
        let mut ans = 0;
        for i in 0..ns {
            if ok[i] {
                let i = uf.find(i);
                ans = std::cmp::max(ans, uf.size[i]);
            }
        }
        println!("{}", ans);
    }
}

struct UnionFind {
    parent: Vec<usize>,
    size: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> UnionFind {
        UnionFind {
            parent: (0..n).collect(),
            size: vec![1; n],
        }
    }

    fn find(&mut self, x: usize) -> usize {
        if self.parent[x] == x {
            x
        } else {
            let p = self.parent[x];
            self.parent[x] = self.find(p);
            self.parent[x]
        }
    }

    fn union(&mut self, x: usize, y: usize) -> bool {
        let mut x = self.find(x);
        let mut y = self.find(y);
        if x == y {
            false
        } else {
            if self.size[x] < self.size[y] {
                std::mem::swap(&mut x, &mut y);
            }
            self.parent[y] = x;
            self.size[x] += self.size[y];
            true
        }
    }
}