#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int mn=2e5+5;
const int mod=998244353;

int n;
char s[mn];
int cal;
ll d[mn][3][3][2];// pos cnt this eq

void spj(){
	bool eq=1;
	for(int i=1;i<n;i++) if(s[i]!=s[i+1]){eq=0;break;}
    if (eq){puts("1");exit(0);}
    if (n==2){
		puts("2");
		exit(0);
    }
    if (n==3){
		int cnt=0;
		if (s[1]==s[3]) cnt++;
		if (s[2]==s[3]) cnt++;
		if (s[1]==s[2]) cnt++;
		if (cnt==0) puts("3");
		else if (s[1]==s[3]&&s[1]!=s[2]&&s[2]!=s[3]) puts("7");
		else puts("6");
		exit(0);
    }

}

int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	bool flag=1;
	for(int i=1;i<=n;i++){
		cal=(cal+s[i]-'a')%3;
		if (s[i]==s[i+1]) flag=0;
	}
	spj();
	d[1][0][0][0]=1;
	for(int k=1;k<=n;k++){//position
		for(int i=0;i<=2;i++){//state
			for(int j=0;j<=2;j++){//previous letter
				if(k==1&&j!=0) continue;
				for(int l=0;l<=2;l++){//nxt letter
					if (j==l&&k!=1){
						d[k+1][(i+l)%3][l][1]+=d[k][i][j][0]+d[k][i][j][1];
						d[k+1][(i+l)%3][l][1]%=mod;
					}
					else{
						d[k+1][(i+l)%3][l][1]+=d[k][i][j][1];
						d[k+1][(i+l)%3][l][1]%=mod;
						d[k+1][(i+l)%3][l][0]+=d[k][i][j][0];
						d[k+1][(i+l)%3][l][0]%=mod;
					}
				}
			}
		}
	}
	printf("%lld\n",(flag+d[n+1][cal][0][1]+d[n+1][cal][1][1]+d[n+1][cal][2][1])%mod);

	return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:

* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).

Constraints

* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.

Input

Input is given from Standard Input in the following format:


S


Output

Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.

Examples

Input

abc


Output

3


Input

abbac


Output

65


Input

babacabac


Output

6310


Input

ababacbcacbacacbcbbcbbacbaccacbacbacba


Output

148010497

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int mn=2e5+5;
const int mod=998244353;

int n;
char s[mn];
int cal;
ll d[mn][3][3][2];// pos cnt this eq

void spj(){
	bool eq=1;
	for(int i=1;i<n;i++) if(s[i]!=s[i+1]){eq=0;break;}
    if (eq){puts("1");exit(0);}
    if (n==2){
		puts("2");
		exit(0);
    }
    if (n==3){
		int cnt=0;
		if (s[1]==s[3]) cnt++;
		if (s[2]==s[3]) cnt++;
		if (s[1]==s[2]) cnt++;
		if (cnt==0) puts("3");
		else if (s[1]==s[3]&&s[1]!=s[2]&&s[2]!=s[3]) puts("7");
		else puts("6");
		exit(0);
    }

}

int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	bool flag=1;
	for(int i=1;i<=n;i++){
		cal=(cal+s[i]-'a')%3;
		if (s[i]==s[i+1]) flag=0;
	}
	spj();
	d[1][0][0][0]=1;
	for(int k=1;k<=n;k++){//position
		for(int i=0;i<=2;i++){//state
			for(int j=0;j<=2;j++){//previous letter
				if(k==1&&j!=0) continue;
				for(int l=0;l<=2;l++){//nxt letter
					if (j==l&&k!=1){
						d[k+1][(i+l)%3][l][1]+=d[k][i][j][0]+d[k][i][j][1];
						d[k+1][(i+l)%3][l][1]%=mod;
					}
					else{
						d[k+1][(i+l)%3][l][1]+=d[k][i][j][1];
						d[k+1][(i+l)%3][l][1]%=mod;
						d[k+1][(i+l)%3][l][0]+=d[k][i][j][0];
						d[k+1][(i+l)%3][l][0]%=mod;
					}
				}
			}
		}
	}
	printf("%lld\n",(flag+d[n+1][cal][0][1]+d[n+1][cal][1][1]+d[n+1][cal][2][1])%mod);

	return 0;
}

```