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 PYTHON3):

Solution:
import sys
from functools import reduce
from collections import Counter
import time
import datetime
import math 

# def time_t():
#     print("Current date and time: " , datetime.datetime.now())
#     print("Current year: ", datetime.date.today().strftime("%Y"))
#     print("Month of year: ", datetime.date.today().strftime("%B"))
#     print("Week number of the year: ", datetime.date.today().strftime("%W"))
#     print("Weekday of the week: ", datetime.date.today().strftime("%w"))
#     print("Day of year: ", datetime.date.today().strftime("%j"))
#     print("Day of the month : ", datetime.date.today().strftime("%d"))
#     print("Day of week: ", datetime.date.today().strftime("%A"))

def ip(): return int(sys.stdin.readline())

# def sip(): return sys.stdin.readline()
def sip() : return input()

def mip(): return map(int,sys.stdin.readline().split())

def mips(): return map(str,sys.stdin.readline().split())

def lip(): return list(map(int,sys.stdin.readline().split()))

def slip(): return list(map(str,sys.stdin.readline().split()))

def matip(n,m):
    lst=[]
    for i in range(n):
        arr = lip()
        lst.insert(i,arr)
    return lst

def factors(n): # find the factors of a number
    return list(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

def minJumps(arr, n): #to reach from 0 to n-1 in the array in minimum steps
    jumps = [0 for i in range(n)]
    if (n == 0) or (arr[0] == 0):
        return float('inf')
    jumps[0] = 0
    for i in range(1, n):
        jumps[i] = float('inf')
        for j in range(i):
            if (i <= j + arr[j]) and (jumps[j] != float('inf')):
                jumps[i] = min(jumps[i], jumps[j] + 1)
                break
    return jumps[n-1]

def dic(arr): # converting list into dict of count
    return Counter(arr)

def check_prime(n):
    if n<2:
        return False
    for i in range(2,int(n**(0.5))+1,2):
        if n%i==0:
            return False
    return True 

# --------------------------------------------------------- #
# sys.stdin = open('input.txt','r')
# sys.stdout = open('output.txt','w')
# --------------------------------------------------------- #

n = ip()
s = sip()
new = ''
flag = 0
count = 0
val = 0
for i in range(n):
    if i%2==0:
        if s[i]=='a' and s[i+1]=='b':
            new += 'ab'
        elif s[i]=='b' and s[i+1]=='a':
            new += 'ba'
        elif s[i]=='b' and s[i+1]=='b':
            count+=1
            new += 'ab'
        elif s[i]=='a' and s[i+1]=='a':
            new += 'ab'
            count+=1
print(count)
print(new)
    



# print(time.process_time())