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)

macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut iter = $ s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; ( iter = $ iter : ident , $ ( $ r : tt ) * ) => { let s = { use std :: io :: Read ; let mut s = String :: new ( ) ; std :: io :: stdin ( ) . read_to_string ( & mut s ) . unwrap ( ) ; s } ; let mut $ iter = s . split_whitespace ( ) ; input_inner ! { $ iter , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { let s = { use std :: io :: Read ; let mut s = String :: new ( ) ; std :: io :: stdin ( ) . read_to_string ( & mut s ) . unwrap ( ) ; s } ; let mut iter = s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; }
macro_rules ! input_inner { ( $ iter : expr ) => { } ; ( $ iter : expr , ) => { } ; ( $ iter : expr , mut $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let mut $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , mut $ var : ident $ ( $ r : tt ) * ) => { input_inner ! { $ iter , mut $ var : usize $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident $ ( $ r : tt ) * ) => { input_inner ! { $ iter , $ var : usize $ ( $ r ) * } } ; }
macro_rules ! read_value { ( $ iter : expr , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ iter , $ t ) ) ,* ) } ; ( $ iter : expr , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ iter , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ iter : expr , { chars : $ base : expr } ) => { read_value ! ( $ iter , String ) . chars ( ) . map ( | c | ( c as u8 - $ base as u8 ) as usize ) . collect ::< Vec < usize >> ( ) } ; ( $ iter : expr , { char : $ base : expr } ) => { read_value ! ( $ iter , { chars : $ base } ) [ 0 ] } ; ( $ iter : expr , chars ) => { read_value ! ( $ iter , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ iter : expr , char ) => { read_value ! ( $ iter , chars ) [ 0 ] } ; ( $ iter : expr , usize1 ) => { read_value ! ( $ iter , usize ) - 1 } ; ( $ iter : expr , $ t : ty ) => { $ iter . next ( ) . unwrap ( ) . parse ::<$ t > ( ) . unwrap ( ) } ; }
#[derive(Debug)]
pub struct KnuthMorrisPratt<T: Eq> {
    pattern: Vec<T>,
    table: Vec<usize>,
}
impl<T: Eq> KnuthMorrisPratt<T> {
    pub fn new(pattern: Vec<T>) -> KnuthMorrisPratt<T> {
        let mut table = vec![0; pattern.len() + 1];
        for i in 1..pattern.len() {
            let mut j = table[i - 1];
            while j > 0 && pattern[i] != pattern[j] {
                j = table[j - 1];
            }
            table[i] = j + (pattern[i] == pattern[j]) as usize;
        }
        KnuthMorrisPratt {
            pattern: pattern,
            table: table,
        }
    }
    pub fn search_all(&self, s: &Vec<T>) -> Vec<usize> {
        let mut res = vec![];
        let mut j = 0;
        for i in 0..s.len() {
            while j > 0 && s[i] != self.pattern[j] {
                j = self.table[j - 1];
            }
            if s[i] == self.pattern[j] {
                j += 1;
            }
            if j == self.pattern.len() {
                res.push(i - (j - 1));
                j = self.table[j - 1];
            }
        }
        res
    }
}
fn chmax<T: Ord>(x: &mut T, y: T) -> bool {
    let b = *x < y;
    if b {
        *x = y;
    }
    b
}
fn main() {
    input! { s: chars, t: chars };
    let n = s.len();
    let m = t.len();
    let mut s2 = s.clone();
    while s2.len() < n + m - 1 {
        s2.extend(s.clone());
    }
    let kmp = KnuthMorrisPratt::new(t);
    let mut a = vec![false; n];
    for i in kmp.search_all(&s2).into_iter() {
        a[i % n] = true;
    }
    let mut used = vec![false; n];
    let mut ans = vec![0; n];
    for i in 0..n {
        if used[i] || !a[i] {
            continue;
        }
        used[i] = true;
        let mut j = i;
        let mut cnt = 0;
        while a[j] {
            j = (j + m) % n;
            cnt += 1;
            if used[j] && a[j] {
                let d = ans[j] + cnt;
                chmax(&mut ans[i], d);
                break;
            }
            chmax(&mut ans[j], cnt);
            used[j] = true;
        }
        if i == j {
            println!("{}", -1);
            return;
        }
        chmax(&mut ans[i], cnt);
    }
    println!("{}", ans.into_iter().max().unwrap());
}