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)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let s: String = scan(tokens);
    let t: String = scan(tokens);
    let mut s = s.into_bytes();
    let t = t.into_bytes();
    let ls = s.len();
    let lt = t.len();
    {
        let mut i = 0;
        while s.len() < ls + lt - 1 {
            let b = s[i];
            s.push(b);
            i += 1;
        }
    }
    let z: Vec<usize> = {
        let mut z = vec![0; lt + 1];
        let mut i = 0;
        let mut j = 1;
        while j < lt {
            if t[i] == t[j] {
                i += 1;
                j += 1;
                z[j] = i;
            } else if i == 0 {
                j += 1;
            } else {
                i = z[i];
            }
        }
        z
    };
    let mut f = vec![false; ls];
    let mut i = 0;
    let mut j = 0;
    while i < s.len() {
        if s[i] == t[j] {
            i += 1;
            j += 1;
            if j == t.len() {
                f[i - t.len()] = true;
                j = z[j];
            }
        } else if j == 0 {
            i += 1;
        } else {
            j = z[j];
        }
    }
    let mut visiting = vec![false; ls];
    let mut dp = vec![None; s.len()];
    fn dfs(
        i: usize,
        ls: usize,
        lt: usize,
        f: &[bool],
        dp: &mut [Option<usize>],
        visiting: &mut [bool],
    ) -> bool {
        if visiting[i] {
            return true;
        }
        if dp[i].is_some() {
            return false;
        }
        if f[i] {
            visiting[i] = true;
            if dfs((i + lt) % ls, ls, lt, f, dp, visiting) {
                return true;
            }
            visiting[i] = false;
            dp[i] = Some(dp[(i + lt) % ls].unwrap() + 1);
            return false;
        } else {
            dp[i] = Some(0);
            return false;
        }
    }
    let mut ans = 0;
    for i in 0..ls {
        if dfs(i, ls, lt, &f, &mut dp, &mut visiting) {
            println!("-1");
            return;
        }
        ans = std::cmp::max(ans, dp[i].unwrap());
    }
    println!("{}", ans);
}