You are given a string s, consisting of lowercase English letters, and the integer m.

One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.

Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.

Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < ... < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j, j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.

Then we take any permutation p of the selected indices and form a new string sip1sip2... sipt.

Find the lexicographically smallest string, that can be obtained using this procedure.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).

The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s.

Output

Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.

Examples

Input

3
cbabc


Output

a


Input

2
abcab


Output

aab


Input

3
bcabcbaccba


Output

aaabb

Note

In the first sample, one can choose the subsequence {3} and form a string "a".

In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab".
import java.util.*;
import java.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=1000000007;
	void work() {
		int m=in.nextInt();
		String str=in.next();
		int n=str.length();
		StringBuilder sb=new StringBuilder();
		PriorityQueue<Integer> pq=new PriorityQueue<>(new Comparator<Integer>() {
			public int compare(Integer a,Integer b) {
				if(str.charAt(a)==str.charAt(b)) {
					return b-a;
				}else {
					return str.charAt(a)-str.charAt(b);
				}
			}
		});
		int i=0;
		for(;i<m;i++) {
			pq.add(i);
		}
		int c=-1;
		while(pq.size()>0) {
			int index=pq.poll();
			if(sb.length()>0&&i==n&&str.charAt(index)-'a'>=c)break;
			c=Math.max(str.charAt(index)-'a', c);
			sb.append(str.charAt(index));
			for(;i<n&&i<=index+m;i++) {
				pq.add(i);
			}
		}
		char[] chs=sb.toString().toCharArray();
		Arrays.sort(chs);
		out.println(new String(chs));
	}
}	



class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}
	
	
	public String next() 
	{
		if(st==null || !st.hasMoreElements())
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

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

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}