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:
T_ON = 0
DEBUG_ON = 0
MOD = 998244353


def solve():
    n = read_int()
    A = list(input())
    count = 0
    for i in range(1, n, 2):
        if A[i] == A[i-1]:
            A[i], A[i-1] = 'a', 'b'
            count += 1
    print(count)
    print("".join(A))


def main():
    T = read_int() if T_ON else 1
    for i in range(T):
        solve()


def debug(*xargs):
    if DEBUG_ON:
        print(*xargs)


from collections import *
import math


#---------------------------------FAST_IO---------------------------------------

import os
import sys
from io import BytesIO, IOBase
# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

#----------------------------------IO_WRAP--------------------------------------

def read_int():
    return int(input())


def read_ints():
    return list(map(int, input().split()))


def print_nums(nums):
    print(" ".join(map(str, nums)))


def YES():
    print("YES")


def Yes():
    print("Yes")


def NO():
    print("NO")


def No():
    print("No")


def First():
    print("First")


def Second():
    print("Second")


#----------------------------------FIB--------------------------------------

def fib(n):
    a, b = 0, 1
    for _ in range(n):
        a, b = b, a + b
    return a


def fib_ns(n):
    assert n >= 1
    f = [0 for _ in range(n + 1)]
    f[0] = 0
    f[1] = 1
    for i in range(2, n + 1):
        f[i] = f[i - 1] + f[i - 2]
    return f

#----------------------------------MOD--------------------------------------

def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


def xgcd(a, b):
    """return (g, x, y) such that a*x + b*y = g = gcd(a, b)"""
    x0, x1, y0, y1 = 0, 1, 1, 0
    while a != 0:
        (q, a), b = divmod(b, a), a
        y0, y1 = y1, y0 - q * y1
        x0, x1 = x1, x0 - q * x1
    return b, x0, y0


def lcm(a, b):
    d = gcd(a, b)
    return a * b // d


def modinv(a, m):
    """return x such that (a * x) % m == 1"""
    g, x, _ = xgcd(a, m)
    if g != 1:
        raise Exception('gcd(a, m) != 1')
    return x % m


def mod_add(x, y):
    x += y
    while x >= MOD:
        x -= MOD
    while x < 0:
        x += MOD
    return x


def mod_mul(x, y):
    return (x * y) % MOD


def mod_pow(x, y):
    if y == 0:
        return 1
    if y % 2:
        return mod_mul(x, mod_pow(x, y - 1))
    p = mod_pow(x, y // 2)
    return mod_mul(p, p)


def mod_inv(y):
    return mod_pow(y, MOD - 2)


def mod_div(x, y):
    # y^(-1): Fermat little theorem, MOD is a prime
    return mod_mul(x, mod_inv(y))

#---------------------------------PRIME---------------------------------------

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


def gen_primes(n):
    """
    generate primes of [1..n] using sieve's method
    """
    P = [True for _ in range(n + 1)]
    P[0] = P[1] = False
    for i in range(int(n ** 0.5) + 1):
        if P[i]:
            for j in range(2 * i, n + 1, i):
                P[j] = False
    return P

#---------------------------------MAIN---------------------------------------

main()