You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.

You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).

Find the lexicographically largest string that can be obtained in this way.

Constraints

* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the lexicographically largest string that satisfies the condition.

Examples

Input

3
aababb


Output

abab


Input

3
bbabaa


Output

bbabaa


Input

6
bbbaabbabaaa


Output

bbbabaaa


Input

9
abbbaababaababbaba


Output

bbaababababa
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ESynchronizedSubsequence solver = new ESynchronizedSubsequence();
        solver.solve(1, in, out);
        out.close();
    }

    static class ESynchronizedSubsequence {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            String s = in.nextString();

            int sum = 0;
            List<String> list = new ArrayList<>();
            int prev = 0;
            for (int i = 0; i < 2 * n; i++) {
                sum += s.charAt(i) == 'a' ? -1 : 1;
                if (sum == 0) {
                    list.add(exec(s.substring(prev, i + 1)));
                    prev = i + 1;
                }
            }

            String ans = list.get(list.size() - 1);
            for (int i = list.size() - 2; i >= 0; i--) {
                if (ans.compareTo(list.get(i) + ans) <= 0) {
                    ans = list.get(i) + ans;
                }
            }

            out.println(ans);
        }

        String exec(String s) {
            int n = s.length() / 2;

            int aCount = 0;
            int bCount = 0;
            int[] aIndex = new int[n];
            int[] bIndex = new int[n];
            for (int i = 0; i < 2 * n; i++) {
                if (s.charAt(i) == 'a') {
                    aIndex[aCount] = i;
                    aCount++;
                } else {
                    bIndex[bCount] = i;
                    bCount++;
                }
            }

            int[] pairIndex = new int[2 * n];
            for (int i = 0; i < n; i++) {
                pairIndex[aIndex[i]] = bIndex[i];
                pairIndex[bIndex[i]] = aIndex[i];
            }

            if (s.charAt(0) == 'a') {
                StringBuilder sb = new StringBuilder();
                for (int i = 0; i < 2 * n; i++) {
                    if (s.charAt(i) == 'a') {
                        sb.append("ab");
                        i = pairIndex[i];
                    }
                }
                return sb.toString();
            } else {
                List<String> list = new ArrayList<>();
                for (int i = 1; i <= n; i++) {
                    boolean[] use = new boolean[2 * n];
                    int count = 0;
                    for (int j = 2 * n - 1; j >= 0; j--) {
                        if (s.charAt(j) == 'a') {
                            use[j] = true;
                            use[pairIndex[j]] = true;

                            count++;
                            if (count == i) break;
                        }
                    }
                    StringBuilder sb = new StringBuilder();
                    for (int j = 0; j < 2 * n; j++) {
                        if (use[j]) sb.append(s.charAt(j));
                    }
                    list.add(sb.toString());
                }
                Collections.sort(list);
                return list.get(list.size() - 1);
            }
        }

    }

    static class InputReader {
        private BufferedReader br;
        private StringTokenizer st;

        public InputReader(InputStream inputStream) {
            br = new BufferedReader(new InputStreamReader(inputStream));
            st = new StringTokenizer("");
        }

        public String nextString() {
            while (!st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine(), " ");
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
            }
            return st.nextToken();
        }

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

    }
}

