Problem: Nikolay got a string s of even length n, which consists only of lowercase Latin letters 'a' and 'b'. Its positions are numbered from 1 to n.

He wants to modify his string so that every its prefix of even length has an equal amount of letters 'a' and 'b'. To achieve that, Nikolay can perform the following operation arbitrary number of times (possibly, zero): choose some position in his string and replace the letter on this position with the other letter (i.e. replace 'a' with 'b' or replace 'b' with 'a'). Nikolay can use no letters except 'a' and 'b'.

The prefix of string s of length l (1 ≤ l ≤ n) is a string s[1..l].

For example, for the string s="abba" there are two prefixes of the even length. The first is s[1...2]="ab" and the second s[1...4]="abba". Both of them have the same number of 'a' and 'b'.

Your task is to calculate the minimum number of operations Nikolay has to perform with the string s to modify it so that every its prefix of even length has an equal amount of letters 'a' and 'b'.

Input

The first line of the input contains one even integer n (2 ≤ n ≤ 2⋅10^{5}) — the length of string s.

The second line of the input contains the string s of length n, which consists only of lowercase Latin letters 'a' and 'b'.

Output

In the first line print the minimum number of operations Nikolay has to perform with the string s to modify it so that every its prefix of even length has an equal amount of letters 'a' and 'b'.

In the second line print the string Nikolay obtains after applying all the operations. If there are multiple answers, you can print any of them.

Examples

Input


4
bbbb


Output


2
abba


Input


6
ababab


Output


0
ababab


Input


2
aa


Output


1
ba

Note

In the first example Nikolay has to perform two operations. For example, he can replace the first 'b' with 'a' and the last 'b' with 'a'. 

In the second example Nikolay doesn't need to do anything because each prefix of an even length of the initial string already contains an equal amount of letters 'a' and 'b'.

Solution (in JAVA):

Solution:
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author wilso
 */
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class codeforces {
    static final long MOD2 = 998_244_353;
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter pw = new PrintWriter(System.out);
       
//        int tc = sc.ni();
        int tc = 1;
        for (int rep = 0; rep < tc; rep++) {
            int N = sc.ni();
            char[] s = sc.next().toCharArray();
            int count = 0;
//            char[] ref = s.toCharArray();
            for (int i = 1; i < N; i+=2) {
                if (s[i-1] == 'a' && s[i] == 'a') {
                    s[i-1] = 'b';
                    count++;
                }
                else if (s[i-1] == 'b' && s[i] == 'b') {
                    s[i-1] = 'a';
                    count++;
                }
            }
            pw.println(count);
            pw.println(new String(s));
        }
        
        pw.close();
    }
 
    
    static String solve(long[] nums) {
      return "a";
        
    }
    
    static int summation(int x) {
        return x * (x+1) / 2;
    }  
    static long pow(long num, long exp, long mod){
        long ans=1;
        for(int i=1;i<=exp;i++){
            ans=(ans*num)%mod;
        }
        return ans;
    }
    static boolean isPrime(int n)
    {
 
        if (n <= 1)
            return false;
        else if (n == 2)
            return true;
        else if (n % 2 == 0)
            return false;
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
    static long gcd(long a, long b) {
        if (a == 0) return b;
        return gcd(b % a, a);
    }
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
    public static void sort(int[] arr) {
        Random rgen = new Random();
        for (int i = 0; i < arr.length; i++) {
            int r = rgen.nextInt(arr.length);
            int temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr);
    }
    public static void sort(long[] arr) {
        Random rgen = new Random();
        for (int i = 0; i < arr.length; i++) {
            int r = rgen.nextInt(arr.length);
            long temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr);
    }
    static int charint(char c) {
        return Integer.parseInt(String.valueOf(c));
    }
    
    /* */
    //printing methods
    /* */
    //WOW!
    /* */
    public static void printArr(PrintWriter pw, int[] arr) {
        StringBuilder sb = new StringBuilder();
        for (int x : arr) {
            sb.append(x + "");
        }
        sb.setLength(sb.length() - 1);
        pw.println(sb.toString());
    }
    public static void printArr2d(PrintWriter pw, int[][] arr) {
        StringBuilder sb = new StringBuilder();
        for (int[] row : arr) {
            for (int x : row) {
                sb.append(x + " ");
            }
            sb.setLength(sb.length() - 1);
            sb.append("\n");
        }
        pw.println(sb.toString());
    }
}

class FastScanner {
    BufferedReader br;
    StringTokenizer st;

    public FastScanner() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
                
            }
        }
        return st.nextToken();
    }

    int ni() {
        return Integer.parseInt(next());
    }

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++)
            ret[i] = ni();
        return ret;
    }

    long nl() {
        return Long.parseLong(next());
    }

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++)
            ret[i] = nl();
        return ret;
    }

    double nd() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}

class MultiSet {
    TreeMap<Long, Integer> map = new TreeMap<>();
    private int size = 0;
    public MultiSet() {
    }
    public void add(Long val) {
        map.put(val, map.getOrDefault(val, 0) + 1);
        size++;
    }
    public void remove(Long val) {
        map.put(val, map.get(val) - 1);
        if (map.get(val) == 0) {
            map.remove(val);
        }
        size--;
    }
    public int size() {
        return size;
    }
    public Long higher(Long val) {
        return map.higherKey(val);
    }
    public Long lower(Long val) {
        return map.lowerKey(val);
    }
    public Long ceiling(Long val) {
        return map.ceilingKey(val);
    }
    public Long floor(Long val) {
        return map.floorKey(val);
    }
    public Long first() {
        return map.firstKey();
    }
    public Long last() {
        return map.lastKey();
    }
    public boolean isEmpty() {
        return map.isEmpty();
    }
}