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: Wrong Answer)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;
use std::num::Wrapping;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

const big_int: Wrapping<u64> = Wrapping(1000_000_000u64 + 7);

fn main() {
    let b_org = read::<String>()
        .chars()
        .map(|x| Wrapping(x as u64))
        .collect::<Vec<_>>();
    let a = read::<String>()
        .chars()
        .map(|x| Wrapping(x as u64))
        .collect::<Vec<_>>();

    let mut b = b_org.clone();
    while b.len() < a.len() * 2 {
        b.append(&mut b_org.clone());
    }

    let mut power_base = Wrapping(1u64);
    for _ in 0..a.len() {
        power_base *= big_int;
    }
    let mut ah = Wrapping(0u64);
    let mut bh = Wrapping(0u64);
    for i in 0..a.len() {
        ah = ah * big_int + a[i];
    }
    for i in 0..a.len() {
        bh = bh * big_int + b[i];
    }

    let mut matched = vec![false; b.len()];
    for i in 0..b.len() {
        matched[i] = ah == bh;
        bh = bh * big_int + b[(i + a.len()) % b.len()] - b[i] * power_base;
    }
    debug!(matched);

    let mut uft = UnionFindTree::new(b_org.len());
    for i in 0..b.len() {
        if matched[i] && matched[(i + a.len()) % b.len()] {
            uft.unite(i % b_org.len(), (i + a.len()) % b_org.len());
        }
    }

    let max_size = (0..b_org.len())
        .map(|x| if matched[x] { uft.get_size(x) } else { 0 })
        .max()
        .unwrap();
    if max_size == b_org.len() {
        println!("{}", -1);
    } else {
        println!("{}", max_size);
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

#[derive(Debug, Clone)]
struct UnionFindTree {
    parent: Vec<isize>,
    size: Vec<usize>,
    height: Vec<u64>,
}

impl UnionFindTree {
    fn new(n: usize) -> UnionFindTree {
        UnionFindTree {
            parent: vec![-1; n],
            size: vec![1usize; n],
            height: vec![0u64; n],
        }
    }

    fn find(&mut self, index: usize) -> usize {
        if self.parent[index] == -1 {
            return index;
        }
        let idx = self.parent[index] as usize;
        let ret = self.find(idx);
        self.parent[index] = ret as isize;
        ret
    }

    fn same(&mut self, x: usize, y: usize) -> bool {
        self.find(x) == self.find(y)
    }

    fn get_size(&mut self, x: usize) -> usize {
        let idx = self.find(x);
        self.size[idx]
    }

    fn unite(&mut self, index0: usize, index1: usize) -> bool {
        let a = self.find(index0);
        let b = self.find(index1);
        if a == b {
            false
        } else {
            if self.height[a] > self.height[b] {
                self.parent[b] = a as isize;
                self.size[a] += self.size[b];
            } else if self.height[a] < self.height[b] {
                self.parent[a] = b as isize;
                self.size[b] += self.size[a];
            } else {
                self.parent[b] = a as isize;
                self.size[a] += self.size[b];
                self.height[a] += 1;
            }
            true
        }
    }
}