You have a garland consisting of nn lamps. Each lamp is colored red, green or blue. The color of the ii-th lamp is sisi ('R', 'G' and 'B' — colors of lamps in the garland).

You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is diverse.

A garland is called diverse if any two adjacent (consecutive) lamps (i. e. such lamps that the distance between their positions is 11) have distinct colors.

In other words, if the obtained garland is tt then for each ii from 11 to n−1n−1 the condition ti≠ti+1ti≠ti+1 should be satisfied.

Among all ways to recolor the initial garland to make it diverse you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.

Input

The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of lamps.

The second line of the input contains the string ss consisting of nn characters 'R', 'G' and 'B' — colors of lamps in the garland.

Output

In the first line of the output print one integer rr — the minimum number of recolors needed to obtain a diverse garland from the given one.

In the second line of the output print one string tt of length nn — a diverse garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.

Examples

input

Copy

9
RBGRRBRGG

output

Copy

2
RBGRGBRGR

input

Copy

8
BBBGBRRR

output

Copy

2
BRBGBRGR

input

Copy

13
BBRRRRGGGGGRR

output

Copy

6
BGRBRBGBGBGRG

思路：根据中间的改

代码；

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

char a[200005];
int main()
{
	int n;
	cin>>n;
	scanf("%s",a);
	int s=0;
	for(int t=0;t<n-1;t++)
	{
		if(a[t]==a[t+1])
		{
			s++;
			if(a[t]!='R'&&a[t+2]!='R')
		    {
		    	a[t+1]='R';
			}
			if(a[t]!='G'&&a[t+2]!='G')
		    {
		    	a[t+1]='G';
			}
			if(a[t]!='B'&&a[t+2]!='B')
		    {
		    	a[t+1]='B';
			}
		}
	
	}
	
		cout<<s<<endl;
		puts(a);
	
	return 0;
}