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: Accepted)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

/*
 * Z algorithm. Calculates an array a[i] = |lcp(s, s[i...|s|])|,
 * where s is the given string.
 * If n = s.length(), the returned array has length n + 1.
 * E.g. z_algorithm("ababa") = {5, 0, 3, 0, 1, 0}
 * Reference: http://snuke.hatenablog.com/entry/2014/12/03/214243
 * Verified by: AtCoder ARC055-C (http://arc055.contest.atcoder.jp/submissions/1061771)
 */
fn z_algorithm(s: &[char]) -> Vec<usize> {
    let n = s.len();
    let mut ret = vec![0; n + 1];
    ret[0] = n;
    let mut i = 1; let mut j = 0;
    while i < n {
        while i + j < n && s[j] == s[i + j] { j += 1; }
        ret[i] = j;
        if j == 0 { i += 1; continue; }
        let mut k = 1;
        while i + k < n && k + ret[k] < j {
            ret[i + k] = ret[k];
            k += 1;
        }
        i += k; j -= k;
    }
    ret
}

fn calc(s: &[char], t: &[char], rep: usize) -> usize {
    let m = s.len();
    let n = t.len();
    let len = rep * max(m, n);
    let mut s_rep = vec![' '; len];
    let mut t_rep = vec![' '; len / n * n];
    for i in 0..len {
        s_rep[i] = s[i % m];
    }
    for i in 0..t_rep.len() {
        t_rep[i] = t[i % n];
    }
    let mut ts = t_rep.clone();
    ts.extend_from_slice(&s_rep);
    let za = z_algorithm(&ts);
    let mut ma = 0;
    for i in t_rep.len()..ts.len() {
        ma = max(za[i], ma);
    }
    min(ma, t_rep.len()) / n
}

fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (write!(out,$($format)*).unwrap());
    }
    input! {
        s: chars,
        t: chars,
    }
    let ma2 = calc(&s, &t, 3);
    let ma3 = calc(&s, &t, 4);
    if ma2 != ma3 {
        puts!("-1\n");
    } else {
        puts!("{}\n", ma2);
    }
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}