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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.*;
import java.util.*;

public class Main implements Runnable {
	public static void main (String[] args) {new Thread(null, new Main(), "_cf", 1 << 28).start();}
	
	public void run() {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		System.err.println("");

		String s = fs.next();
		String t = fs.next();
		StringBuilder ns = new StringBuilder();
		ns.append(s);
		while(ns.length() < 2*t.length()+s.length()) ns.append(s);
		int v1 = runs(t, ns.toString());
		ns.append(ns.toString());
		int v2 = runs(t, ns.toString());
		
		if(v2 > v1) v1 = -1;
		
		out.println(v1);
		out.close();
	}
	
	int runs(String t, String s) {
		int[] z = zValues(t+"#"+s);
		int max = 0;
		
		z = Arrays.copyOfRange(z, t.length()+1, z.length);
		for(int i = 0; i < z.length; i++) {
			int runs = 0;
			int sub = 0;
			while(i < z.length && z[i] >= t.length()) {
				runs++;
				sub = 1;
				i += t.length();
			}
			i -= sub;
			max = Math.max(max, runs);
		}
		
		return max;
	}
	
	int[] zValues(String string) {
		int n = string.length();  char[] s = string.toCharArray();
		int[] z = new int[n];  z[0] = n;
		int L = 0, R = 0;
		int[] left = new int[n], right = new int[n];
		for(int i = 1; i < n; ++i) {
			if(i > R) {
				L = R = i;
				while(R < n && s[R-L] == s[R]) R++;
				z[i] = R - L;  R--;
			} else {
				int k = i-L;
				if(z[k] < R-i+1) z[i] = z[k];
				else {
					L = i;
					while(R < n && s[R-L] == s[R]) R++;
					z[i] = R - L;  R--;
				}
			}
			left[i] = L;  right[i] = R;
		}
		return z;
	}
	
	class FastScanner {
		public int BS = 1<<16;
		public char NC = (char)0;
		byte[] buf = new byte[BS];
		int bId = 0, size = 0;
		char c = NC;
		double num = 1;
		BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) {
			try {
				in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
			}
			catch (Exception e) {
				in = new BufferedInputStream(System.in, BS);
			}
		}

		public char nextChar(){
			while(bId==size) {
				try {
					size = in.read(buf);
				}catch(Exception e) {
					return NC;
				}                
				if(size==-1)return NC;
				bId=0;
			}
			return (char)buf[bId++];
		}

		public int nextInt() {
			return (int)nextLong();
		}

		public long nextLong() {
			num=1;
			boolean neg = false;
			if(c==NC)c=nextChar();
			for(;(c<'0' || c>'9'); c = nextChar()) {
				if(c=='-')neg=true;
			}
			long res = 0;
			for(; c>='0' && c <='9'; c=nextChar()) {
				res = (res<<3)+(res<<1)+c-'0';
				num*=10;
			}
			return neg?-res:res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c!='.' ? cur:cur+nextLong()/num;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c>32) {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c!='\n') {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if(c>32)return true;
			while(true) {
				c=nextChar();
				if(c==NC)return false;
				else if(c>32)return true;
			}
		}
		
		public int[] nextIntArray(int n) {
			int[] res = new int[n];
			for(int i = 0; i < n; i++) res[i] = nextInt();
			return res;
		}
		
	}

}