Diverse Garland CodeForces - 1108D （贪心+暴力枚举）

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

9
RBGRRBRGG

Output

2
RBGRGBRGR

Input

8
BBBGBRRR

Output

2
BRBGBRGR

Input

13
BBRRRRGGGGGRR

Output

6
BGRBRBGBGBGRG

题意：给定一个字符串，只包含RGB三个字符，你可以改变某些字符使之这个字符串相邻的字符不相等。
那么我们只需要枚举从第二个字符开始的每一个字符串，判定是否和前面的字符相等，如果相等就改成不和后面字串相等字符，这样消耗就一定最小。
很水的一题，细节看code。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
char s[maxn];
int main()
{
    gg(n);
    scanf("%s",s);
    int ans=0;
    repd(i,1,n-1)
    {
        if(s[i]==s[i-1])
        {
            if(s[i]=='B')
            {
                if(s[i+1]!='R')
                {
                    s[i]='R';
                }else
                {
                    s[i]='G';
                }
            }else if(s[i]=='R')
            {
                if(s[i+1]!='B')
                {
                    s[i]='B';
                }else
                {
                    s[i]='G';
                }
            }else if(s[i]=='G')
            {

                if(s[i+1]!='R')
                {
                    s[i]='R';
                }else
                {
                    s[i]='B';
                }
            }
            ans++;
        }

    }
    printf("%d\n",ans );
    printf("%s\n", s);
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}