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:
import java.util.*;
import java.io.*;

public class Main {
	static int mod = 1000000007;
 	public static void main(String[] args){
		FastScanner scanner = new FastScanner();
		int n = scanner.nextInt();
		String s = scanner.next();
		int acnt = 0;
		int bcnt = 0;
		char[] ans = new char[n];
		for(int i = 0; i < n; i++){
			ans[i] = s.charAt(i);
		}
		int c = 0;
		for(int i = 0; i < n; i++){
			if(s.charAt(i) == 'a'){
				acnt++;
			}else if(s.charAt(i) == 'b'){
				bcnt++;
			}
			if(i % 2 == 1 && acnt != bcnt){
				c++;
				if(acnt < bcnt){
					ans[i] = 'a';
					acnt++;
					bcnt--;
				}else if(acnt > bcnt){
					ans[i] = 'b';
					bcnt++;
					acnt--;
				}
			}else{
				ans[i] = s.charAt(i);
			}
		}
		System.out.println(c);
		System.out.println(ans);
	}
	static class UnionFind {
		int[] parent;
		public UnionFind(int size) {
			parent = new int[size];
			Arrays.fill(parent, -1);
		}
		public boolean unite(int x, int y) {
			x = root(x);
			y = root(y);
			if (x != y) {
				if (parent[y] < parent[x]) {
					int tmp = y;
					y = x;
					x = tmp;
				}
				parent[x] += parent[y];
				parent[y] = x;
				return true;
			}
			return false;
		}
		public boolean same(int x, int y) {
			return root(x) == root(y);
		}
		public int root(int x) {
			return parent[x] < 0 ? x : (parent[x] = root(parent[x]));
		}
		public int size(int x) {
			return -parent[root(x)];
		}
	}
	static class Pair implements Comparable<Pair>{
    int first, second;
    Pair(int a, int b){
        first = a;
        second = b;
    }
    @Override
    public boolean equals(Object o){
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair p = (Pair) o;
        return first == p.first && second == p.second;
    }
    @Override
    public int compareTo(Pair p){
        //return first == p.first ? second - p.second : first - p.first; //firstで昇順にソート
        //return (first == p.first ? second - p.second : first - p.first) * -1; //firstで降順にソート
        //return second == p.second ? first - p.first : second - p.second;//secondで昇順にソート
        //return (second == p.second ? first - p.first : second - p.second)*-1;//secondで降順にソート
		    //return first * 1.0 / second > p.first * 1.0 / p.second ? 1 : -1; // first/secondの昇順にソート
		    //return first * 1.0 / second < p.first * 1.0 / p.second ? 1 : -1; // first/secondの降順にソート
				//return second * 1.0 / first > p.second * 1.0 / p.first ? 1 : -1; // second/firstの昇順にソート
		    //return second * 1.0 / first < p.second * 1.0 / p.first ? 1 : -1; // second/firstの降順にソート
				//return Math.atan2(second, first) > Math.atan2(p.second, p.first) ? 1 : -1; // second/firstの昇順にソート
				//return first + second > p.first + p.second ? 1 : -1; //first+secondの昇順にソート
				return first + second < p.first + p.second ? 1 : -1; //first+secondの降順にソート
				//return first - second < p.first - p.second ? 1 : -1; //first+secondの降順にソート
		}
  }

	private static class FastScanner {
		private final InputStream in = System.in;
		private final byte[] buffer = new byte[1024];
		private int ptr = 0;
		private int buflen = 0;
		private boolean hasNextByte() {
				if (ptr < buflen) {
						return true;
				}else{
						ptr = 0;
						try {
								buflen = in.read(buffer);
						} catch (IOException e) {
								e.printStackTrace();
						}
						if (buflen <= 0) {
								return false;
						}
				}
				return true;
		}
		private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
		private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
		public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
		public String next() {
				if (!hasNext()) throw new NoSuchElementException();
				StringBuilder sb = new StringBuilder();
				int b = readByte();
				while(isPrintableChar(b)) {
						sb.appendCodePoint(b);
						b = readByte();
				}
				return sb.toString();
		}
		public long nextLong() {
				if (!hasNext()) throw new NoSuchElementException();
				long n = 0;
				boolean minus = false;
				int b = readByte();
				if (b == '-') {
						minus = true;
						b = readByte();
				}
				if (b < '0' || '9' < b) {
						throw new NumberFormatException();
				}
				while(true){
						if ('0' <= b && b <= '9') {
								n *= 10;
								n += b - '0';
						}else if(b == -1 || !isPrintableChar(b)){
								return minus ? -n : n;
						}else{
								throw new NumberFormatException();
						}
						b = readByte();
				}
		}
		public int nextInt() {
				long nl = nextLong();
				if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
				return (int) nl;
		}
		public double nextDouble() { return Double.parseDouble(next());}
	}
}