Problem: Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.

A little later they found a string s, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring t of the string s.

Prefix supposed that the substring t is the beginning of the string s; Suffix supposed that the substring t should be the end of the string s; and Obelix supposed that t should be located somewhere inside the string s, that is, t is neither its beginning, nor its end.

Asterix chose the substring t so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring t aloud, the temple doors opened. 

You know the string s. Find the substring t or determine that such substring does not exist and all that's been written above is just a nice legend.

Input

You are given the string s whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.

Output

Print the string t. If a suitable t string does not exist, then print "Just a legend" without the quotes.

Examples

Input

fixprefixsuffix


Output

fix

Input

abcdabc


Output

Just a legend

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * 类InOutHandler.java的实现描述：TODO 类实现描述
 * kmp 根据搜索词--》获取 部分匹配表 ：http://www.ruanyifeng.com/blog/2013/05/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm.html
 * "部分匹配值"就是"前缀"和"后缀"的最长的共有元素的长度
 * @author zhouneng.zn 2017年12月20日 下午8:24:05
 */
public class MyFindMaxLengthStr {

	public static void main(String[] args) throws Exception {
		// FileInputStream inputStream = new FileInputStream(new
		// File("/Users/yexin/Lesson/ACM/charp_1/sort/src/textE.txt"));
		InputStream inputStream = System.in;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(System.out);
		E e = new E();
		for (int i = 0; i < 1; i++) {
			e.solve(in, out);
			out.flush();
		}
	}

	static class E {
		public void solve(InputReader in, PrintWriter out) {
			String s = in.next();
			int slen = s.length();
			int[] next = getNext(s);

			// exist[*] =true 表示匹配上了  "ABCDABDAB" ==》 -1000012012
			boolean[] exist = new boolean[slen + 1];
			for (int i = 1; i < next.length - 1; i++) {
				exist[next[i]] = true;
			}

			int maxSubStrIndex = next[slen];

			while (maxSubStrIndex != 0) { // next末位为0，说明没有最大公共长度
				if (exist[maxSubStrIndex]) {
					out.println(s.substring(0, maxSubStrIndex));
					break;
				} else {
					maxSubStrIndex = next[maxSubStrIndex];
				}
			}

			if (maxSubStrIndex == 0) {
				out.println("Just a legend");
			}
		}

		private int[] getNext(String s) {
			int slen = s.length();
			int[] next = new int[slen + 1];
			next[0] = -1;
			// k就是前缀和后缀的最大长度
			int k = -1;
			int j = 0;
			while (j < slen) {
				// "ABCDABD"
				if (k == -1 || s.charAt(k) == s.charAt(j)) {
					k++;
					j++;
					next[j] = k;
				} else {
					k = next[k];
				}
			}
//			for (int ii=0;ii<next.length;ii++) {
//				System.out.print(next[ii]);
//			}System.out.println();
			return next;
		}
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 102400);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

	}
}