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.OutputStream;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Template built using CHelper plug-in
 * Actual solution is at the top
 * @author flk
 */
public class main {

    public static void main(String[] args) throws InterruptedException {

        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(1, in, out);
        out.close();

    }
}



class Task {

    public void solve(int testNumber, Scanner in, PrintWriter out) {

        int m = in.nextInt(); in.nextLine();
        String s = in.nextLine();

        int[] lettersAll = new int[26];
        int[] lettersCurrent = new int[26];
        int[] haveto = new int[26];

        //find 'haveto' letters
        int i = 0;
        for (i = 0; i < m; i++) {
            lettersCurrent[s.charAt(i) - 'a']++;
            lettersAll[s.charAt(i) - 'a']++;
        }

        int j = 0;
        while (lettersCurrent[j] == 0) j++;
        haveto[j]++;


        while (i < s.length()) {
            lettersCurrent[s.charAt(i-m) - 'a']--;
            lettersCurrent[s.charAt(i) - 'a']++;
            lettersAll[s.charAt(i) - 'a']++;

            j = 0;
            while (lettersCurrent[j] == 0) j++;
            haveto[j]++;
            i++;
        }

        //grab all letters below max + num of maxes

        StringBuilder sb = new StringBuilder();
        int maxLetter = 25;
        while (haveto[maxLetter] == 0) maxLetter--;

        for (int k = 0; k < maxLetter; k++) {
            for (int l = 0; l < lettersAll[k]; l++) {
                sb.append((char)(k + 'a'));
            }
        }


        //find min of maxLetter
        int start = 0;
        int nummax = 0;
        int end = start + m -1;
        lettersCurrent = new int[26];
        for (i = start; i <= end; i++) {
            lettersCurrent[s.charAt(i) - 'a']++;
        }

        j = 0;
        while (lettersCurrent[j] == 0) j++;

        int k = 0;
        if (j < maxLetter) {
            // find closest small
            while(s.charAt(start + k) != 'a' + j) k++;
            k++;
        }
        else {
            //find last max
            while(s.charAt(end - k) != 'a' + j) k++;
            k = m - k;
            nummax++;
        }

        boolean solution = false;
        while (k > 0) {
            start++; end++;
            if (end == s.length()) {
                solution = true;
                break;
            }
            lettersCurrent[s.charAt(start-1) - 'a']--;
            lettersCurrent[s.charAt(end) - 'a']++;
            k--;
        }

        while (end < s.length()) {
            j = 0;
            while (lettersCurrent[j] == 0) j++;

            k = 0;
            if (j < maxLetter) {
                // find closest small
                while(s.charAt(start + k) != 'a' + j) k++;
                k++;
            }
            else {
                //find last max
                while(s.charAt(end - k) != 'a' + j) k++;
                k = m - k;
                nummax++;
            }

            while (k > 0) {
                start++; end++;
                if (end == s.length()) break;
                lettersCurrent[s.charAt(start-1) - 'a']--;
                lettersCurrent[s.charAt(end) - 'a']++;
                k--;


            }

        }


        for (int l = 0; l < nummax; l++) {
            sb.append((char) (maxLetter + 'a') );
        }

        out.println(sb);

    }

}




