Codeforces Round #587 (Div. 3) A - Prefixes

原文链接：https://www.cnblogs.com/xwl3109377858/p/11564167.html

Codeforces Round #587 (Div. 3)

A - Prefixes

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⋅105) — 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 evenlength 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'.

 

题意：题目意思大概就是给你一个由 'a'、'b' 构成的字符串，问你最少改变几个字符

使这个字符串的每个偶数前缀的 'a'、'b' 字符数量相同，并输出改变后的字符串。

思路：直接扫一遍，每次找两个字符，如果数量不同就改。

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<list>
#include<stack>
using namespace std;
#define ll long long 
const int mod=1e9+7;
const int inf=1e9+7;
 
//const int maxn=
 
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    
    int n;
    cin>>n;
    string str;
    cin>>str;
    
    int cnt_a=0,cnt_b=0;
    
    int ans=0;
    
    for(int i=0;i<n;i++)
    {
        if(str[i]=='a')
            cnt_a++;
        else if(str[i]=='b')
            cnt_b++;
        if(i&1)
        {
            if(cnt_a!=cnt_b)
            {
                ans++;
                if(cnt_a==2)
                    str[i]='b';
                else if(cnt_b==2)
                    str[i]='a';    
            }
            cnt_a=0;
            cnt_b=0;
        }
    }
    
    cout<<ans<<endl;
    cout<<str<<endl;
    
    return 0;
}