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)

use std::io::Read;

fn z_algorithm(s: &Vec<char>) -> Vec<usize> {
    let n = s.len();
    let mut a = vec![0; n];
    a[0] = n;
    let mut i = 1;
    let mut j = 0;
    while i < n {
        while i + j < n && s[j] == s[i + j] {
            j += 1;
        }
        a[i] = j;
        if j == 0 {
            i += 1;
            continue;
        }
        let mut k = 1;
        while i + k < n && k + a[k] < j {
            a[i + k] = a[k];
            k += 1;
        }
        i += k;
        j -= k;
    }
    a
}

fn calc(x: usize, state: &mut [usize], dp: &mut [usize], z: &[usize], n: usize, m: usize) -> usize {
    if state[x] == 1 {
        return std::usize::MAX / 2;
    }
    if state[x] == 2 {
        return dp[x];
    }
    state[x] = 1;
    dp[x] = if z[x] >= m {
        let next = (x + m) % n;
        calc(next, state, dp, z, n, m) + 1
    } else {
        0
    };
    state[x] = 2;
    return dp[x];
}

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let s: Vec<char> = it.next().unwrap().chars().collect();
    let t: Vec<char> = it.next().unwrap().chars().collect();
    let mut a = vec![];
    while a.len() < 2 * std::cmp::max(s.len(), t.len()) {
        a.append(&mut s.clone());
    }
    let mut b = t.clone();
    b.push('#');
    b.append(&mut a.clone());
    let z = z_algorithm(&b).split_off(t.len() + 1);
    let n = s.len();
    let m = t.len();
    let mut state = vec![0; n];
    let mut dp = vec![0; n];
    for i in 0..s.len() {
        if state[i] == 2 {
            continue;
        }
        if std::usize::MAX / 2 <= calc(i, &mut state, &mut dp, &z, n, m) {
            println!("-1");
            return;
        }
    }
    let ans = *dp.iter().max().unwrap();
    println!("{}", ans);
}

fn main() {
    run();
}