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_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($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)*}
    };
}

#[allow(unused_macros)]
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)*}
    };

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

#[allow(unused_macros)]
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:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*);
    }
}

#[allow(unused_imports)]
use std::cmp::{min, max};
#[allow(unused_imports)]
use std::io::{stdout, stdin, BufWriter, Write};

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

#[derive(Debug)]
pub struct UnionFind {
    par: Vec<usize>,
    size: Vec<usize>,
}
impl UnionFind {
    pub fn new(n: usize) -> UnionFind {
        let mut vec = vec![0; n];
        for i in 0..n {
            vec[i] = i;
        }
        UnionFind { par: vec, size: vec![1; n] }
    }
    pub fn find(&mut self, x: usize) -> usize {
        if x == self.par[x] {
            x
        } else {
            let par = self.par[x];
            let res = self.find(par);
            self.par[x] = res;
            res
        }
    }
    pub fn same(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }
    pub fn unite(&mut self, a: usize, b: usize) {
        let apar = self.find(a);
        let bpar = self.find(b);
        if apar != bpar {
            self.size[bpar] += self.size[apar];
        }
        self.par[apar] = bpar;
    }
}

fn main() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (write!(out,$($format)*).unwrap());
    }

    input!{
      s: String,
      t: String,
    }
    let mut ss = s.clone();
    ss += s.as_str();
    while ss.len() < t.len() * 2 {
        ss += s.as_str();
    }
    let temp: Vec<_> = (t.clone() + "$" + ss.as_str()).chars().collect();
    let z_array = z_algorithm(&temp);

    let n = s.len();
    let mut uf = UnionFind::new(n);
    for i in 0..n {
        let p = z_array[ i + t.len() + 1];
        if p == t.len() {
            let nx = (i+t.len())%n;
            if uf.same(i, nx) {
                return puts!("{}\n", -1);
            }
            uf.unite(i, nx);
        }
    }

    let mut ans = 1;
    for i in 0..n {
        ans = max(ans, uf.size[i]);
    }
    puts!("{}\n", ans-1);
}