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 CPP):

Solution:
#include <bits/stdc++.h>
const long long int mod = 9999999999999983;
long long int primes[6] = {1125899906842597, 1495921043, 1005985879,
                           1495921043,       1005985879, 1495921043};
using namespace std;
vector<long long int> adj[2000009];
long long int parent[2000009];
long long int vis[2000009];
long long int level[2000009];
long long int dist[2000009];
long long int dp[100009];
long long int arr[100009];
long long int brr[2000009];
long long int crr[2000009];
long long int hashing[2000009];
long long int ar[509][509];
long long int br[509][509];
long long int cr[509][509];
long long int multiply(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int add(long long int a, long long int b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long int sub(long long int a, long long int b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = ((res % p) * (x % p)) % p;
    y = y >> 1;
    x = ((x % p) * (x % p)) % p;
  }
  return res;
}
int main() {
  int start_s = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  vector<long long int> vect, vt, vct;
  vector<long long int>::iterator itt;
  set<long long int> st;
  set<long long int>::iterator itr;
  map<long long int, long long int> mp, mpp, mp1, mp2;
  map<long long int, long long int>::iterator it;
  long long int i, j, k, n, m, p, res1, res2, q, t, a, b, c,
      maxi = INT_MIN, mini = INT_MAX, sum = 0, ans = 0, res = 0, val = 0,
      ans1 = 0, ans2 = 0, rem = 0, diff = 0, cnt = 0, l, r, flag = 0, e, index,
      val1 = 0, val2 = 0, z, h = 0, u, v, mid, len, tot, fl = 0;
  string str, str1, str2;
  cin >> n;
  cin >> str;
  for (i = 0; i < str.size(); i += 2) {
    if (str[i] == 'a' && str[i + 1] == 'a') {
      str[i + 1] = 'b';
      cnt++;
    }
    if (str[i] == 'b' && str[i + 1] == 'b') {
      str[i + 1] = 'a';
      cnt++;
    }
  }
  cout << cnt << endl;
  cout << str;
}