Problem: Alice bought a Congo Prime Video subscription and was watching a documentary on the archaeological findings from Factor's Island on Loch Katrine in Scotland. The archaeologists found a book whose age and origin are unknown. Perhaps Alice can make some sense of it?

The book contains a single string of characters "a", "b" and "c". It has been pointed out that no two consecutive characters are the same. It has also been conjectured that the string contains an unusually long subsequence that reads the same from both sides. 

Help Alice verify this by finding such subsequence that contains at least half of the characters of the original string, rounded down. Note that you don't have to maximise the length of it.

A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.

Input

The input consists of a single string s (2 ≤ |s| ≤ 10^6). The string s consists only of characters "a", "b", "c". It is guaranteed that no two consecutive characters are equal.

Output

Output a palindrome t that is a subsequence of s and |t| ≥ ⌊ (|s|)/(2) ⌋.

If there are multiple solutions, you may print any of them. You don't have to maximise the length of t.

If there are no solutions, output a string "IMPOSSIBLE" (quotes for clarity).

Examples

Input


cacbac


Output


aba


Input


abc


Output


a


Input


cbacacacbcbababacbcb


Output


cbaaacbcaaabc

Note

In the first example, other valid answers include "cacac", "caac", "aca" and "ccc". 

Solution (in JAVA):

Solution:
import java.io.*; 
import java.math.*;
import java.util.*;
 
public class Main {
	static final long MOD = 998244353;
	//static final long MOD = 1000000007;
	static boolean[] visited;
    public static void main(String[] args) throws IOException {
        FastScanner sc = new FastScanner();
        String s = sc.next();
        int N = s.length();
        int L = 0;
        int H = N-1;
        StringBuilder ans = new StringBuilder();
        while (L <= H-3) {
        	int[] cnt = new int[3];
        	cnt[s.charAt(L)-'a'] += 1;
        	cnt[s.charAt(L+1)-'a'] += 1;
        	cnt[s.charAt(H)-'a'] += 1;
        	cnt[s.charAt(H-1)-'a'] += 1;
        	if (cnt[0] > 1) {
        		ans.append('a');
        	} else if (cnt[1] > 1) {
        		ans.append('b');
        	} else {
        		ans.append('c');
        	}
        	L += 2;
        	H -= 2;
        }
    	System.out.print(ans);
        if (L == H-1 || L == H-2)
        	System.out.print(s.charAt(L));
    	System.out.print(ans.reverse());
    }
    
    public static long power(long x, long y, long p) { 
        // Initialize result
        long res = 1;
         
        // Update x if it is more   
        // than or equal to p 
        x = x % p;  
      
        while (y > 0) 
        { 
            // If y is odd, multiply x 
            // with result 
            if((y & 1)==1) 
                res = (res * x) % p; 
      
            // y must be even now 
            // y = y / 2 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    }
    
    public static long dist(int[] point, int[] point2) {
    	return (long)(Math.pow((point2[1]-point[1]),2)+Math.pow((point2[0]-point[0]),2));
    }
    
    public static long gcd(long a, long b) {
    	if (b == 0)
    		return a;
    	else
    		return gcd(b,a%b);
    }
    
    public static int[][] sort(int[][] array) {
    	//Sort an array (immune to quicksort TLE)
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    int[] temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		Arrays.sort(array, new Comparator<int[]>() {
			  @Override
        	  public int compare(int[] arr1, int[] arr2) {
				  return arr1[0]-arr2[0];
	          }
		});
		return array;
	}
    
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next() { 
            while (st == null || !st.hasMoreElements()) { 
                try { 
                    st = new StringTokenizer(br.readLine());
                } catch (IOException  e) { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() { 
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str; 
        }
    }
}
class Node {
	public HashSet<Node> children;
	public int n;
	
	public Node(int n) {
		this.n = n;
		children = new HashSet<Node>();
	}
	
	public void addChild(Node node) {
		children.add(node);
	}
	
	public void removeChild(Node node) {
		children.remove(node);
	}

	/* (non-Javadoc)
	 * @see java.lang.Object#hashCode()
	 */
	@Override
	public int hashCode() {
		return n;
	}

	@Override
	public boolean equals(Object obj) {
		if (! (obj instanceof Node)) {
			return false;
		} else {
			Node node = (Node) obj;
			return (n == node.n);
		}
	}
}