And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output

Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

Examples

Input

3 3
abc


Output

aca


Input

3 2
abc


Output

ac


Input

3 3
ayy


Output

yaa


Input

2 3
ba


Output

baa

Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.
import java.util.Scanner;
import java.util.Set;
import java.util.SortedSet;
import java.util.TreeSet;

public class C {

	static SortedSet<Integer> valueSet = new TreeSet<>();
	private static int MAX_VALUE;
	private static int MIN_VALUE;
	
	public static void main(String[] args) throws InterruptedException {
		
		Integer length;
		String t;
		
		Scanner sc = new Scanner(System.in);
		
		sc.nextInt();
		length = sc.nextInt();
		
		t = sc.next(); 
		
		//t= "fjrhrlxvhpljltpnfxpvjttjhfrxxrnrlhrxhldxhbvbvjbrflnfnbdjxflpvnplrrfttrpfrjlnbvtldfhfvjfxbhpbplnprhdppldprjfxvpxjpjttbrnvbndnhnbfnblxttxfvvnnllfhlntpxthtbdjhvjltrbhlxblvfxdxjlfjvhttpfvrtlxtxbldrtfhjlxrbblpbnxbfdrtvrrffhvpxfbndbppdbbjlhjjllthtxhtfvxrhtfvfltjxttjtvfthlvtplhlhjddrdjnbjnthtnpntbpdpxdbpnpdbhlrfhphjpbtlpjprdvhnlvlndvxpbbrfbttrrdbbjvxvjpnhbfbthxrrnljxthhjlxjnbvxdvprlthpblhphpvxjhdxlrbxlhhhrrpxvxbvrpnhxvdxnrlvnjbjxdpbnnpdfbnthvdjldxdptxjjnnjbtrbxdtthhdvjflltfnfxjhhtpnnxfjfhnndhpdhxbhxf";
		//length = 200;
		
		String solution = "";

		for(int i=0; i<t.length(); i++){
			//System.out.println(5);
			valueSet.add((int)(t.charAt(i)));
		}
		
		//System.out.println(valueAndChar.keySet().size());
		
		MAX_VALUE = valueSet.last();
		MIN_VALUE = valueSet.first();
		
		StringBuffer strB = new StringBuffer();
		
		if(length<=t.length()){
			int i=length-1;
				
				//the tail, should be small
				while(t.charAt(i) == MAX_VALUE){
					//System.out.println("adding "+minValue(valueAndChar.keySet()));
					strB.insert(0,(char)MIN_VALUE);
					//strB.insert(0,solution);
					i--;
				}
				
				//add the next character 
				int nextChar = t.charAt(i) + 1;
				while(!valueSet.contains(nextChar)){
					nextChar++;
					//System.out.println(2);
				}
				
				strB.insert(0,(char)nextChar);
				
				//the head, should be the same
				
				while(i>0){
					strB.insert(0,t.charAt(i-1));
					//strB.insert(0,solution);
					i--;
				}
				

			System.out.println(strB.toString());
			
		} else {
			strB.insert(0,t);
			
			while(strB.length()<length){
				strB.append((char)MIN_VALUE);
				//System.out.println(4);
			}
			System.out.println(strB.toString());
		}
		
		
	}
	
	static int maxValue(Set<Integer> set){
		return set.stream().max((i1,i2)->{
			return i1-i2;
		}).get();
	}
	
	static int minValue(Set<Integer> set){
		return set.stream().min((i1,i2)->{
			return i1-i2;
		}).get();
	}
	
}
