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::*;
use std::str::FromStr;

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

const MODS: [i64; 3] = [1000000007i64, 1000000009i64, 37582157i64];
const BASES: [i64; 3] = [37582157i64, 1000000007i64, 997i64];
fn ok(cnt: usize, s: &Vec<char>, t: &Vec<char>) -> bool {
    'outer_loop: for _iter in 0..3 {
        let m = MODS[_iter];
        let b = BASES[_iter];
        let mut target_hash = 0;
        let mut high = 1;
        for _ in 0..cnt {
            for i in 0..t.len() {
                target_hash = (target_hash * b + t[i] as u8 as i64) % m;
                high = (high * b) % m;
            }
        }
        let len = cnt * t.len();
        let target_hash = target_hash;
        let mut used = 0;
        let mut hash = 0;
        loop {
            for i in 0..s.len() {
                hash = (hash * b + s[i] as u8 as i64) % m;
                used += 1;
                if used > len {
                    let pos = (i + s.len() * len - len) % s.len();
                    hash = (hash - s[pos] as u8 as i64 * high + m * m) % m;
                }
                if used >= len && hash == target_hash {
                    continue 'outer_loop;
                }
            }
            if used >= len * 2 && used >= s.len() * 3 {
                break;
            }
        }
        return false;
    }
    return true;
}

fn main() {
    let s = read::<String>().chars().collect::<Vec<char>>();
    let t = read::<String>().chars().collect::<Vec<char>>();
    let mut l = 0;
    let mut r = s.len() * 3 / t.len() + 3;
    if ok(r, &s, &t) {
        println!("-1");
        return;
    }
    while r - l > 1 {
        let mid = (l + r) / 2;
        if ok(mid, &s, &t) {
            l = mid;
        } else {
            r = mid;
        }
    }
    println!("{}", l);
}