#include<bits/stdc++.h>
using namespace std;
const int N=200005,mod=998244353;
char s[N];
int f[N][3][3];
int power(int x,int y)
{
	int s=1;
	for(;y;y/=2,x=1ll*x*x%mod)
		if(y&1) 
			s=1ll*s*x%mod;
	return s;
}
int main()
{
	scanf("%s",s+1);
	int n=strlen(s+1);
	if(n==2)
	{
		printf("%d\n",1+(s[1]!=s[2]));
		return 0;
	}
	if(n==3)
	{
		if(s[1]==s[2]&&s[2]==s[3]) 
			puts("1");
		else if(s[1]==s[2]||s[2]==s[3]) 
			puts("6");
		else if(s[1]==s[3]) 
			puts("7");
		else 
			puts("3");
		return 0;
	}
	int fl=1;
	for(int i=1;i<n;i++)
		if(s[i]!=s[i+1]) 
			fl=0;
	if(fl)
	{
		puts("1");
		return 0;
	}
	int ans=power(3,n-1);
	int v=0;
	for(int i=1;i<=n;i++)
		v=(v+s[i]-'a')%3;
	for(int i=0;i<3;i++)
		f[1][i][i]=1;
	for(int i=2;i<=n;i++)
		for(int j=0;j<3;j++)
			for(int k=0;k<3;k++)
				for(int l=0;l<3;l++)
					if(k!=l)
						f[i][(j+l)%3][l]=(f[i][(j+l)%3][l]+f[i-1][j][k])%mod;
	for(int i=0;i<3;i++)
		ans=(ans+mod-f[n][v][i])%mod;
	fl=1;
	for(int i=1;i<n;i++)
		if(s[i]==s[i+1]) 
			fl=0;
	printf("%d\n",(ans+fl)%mod);
	return 0;
}

### Prompt

Create a solution in cpp for 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;
const int N=200005,mod=998244353;
char s[N];
int f[N][3][3];
int power(int x,int y)
{
	int s=1;
	for(;y;y/=2,x=1ll*x*x%mod)
		if(y&1) 
			s=1ll*s*x%mod;
	return s;
}
int main()
{
	scanf("%s",s+1);
	int n=strlen(s+1);
	if(n==2)
	{
		printf("%d\n",1+(s[1]!=s[2]));
		return 0;
	}
	if(n==3)
	{
		if(s[1]==s[2]&&s[2]==s[3]) 
			puts("1");
		else if(s[1]==s[2]||s[2]==s[3]) 
			puts("6");
		else if(s[1]==s[3]) 
			puts("7");
		else 
			puts("3");
		return 0;
	}
	int fl=1;
	for(int i=1;i<n;i++)
		if(s[i]!=s[i+1]) 
			fl=0;
	if(fl)
	{
		puts("1");
		return 0;
	}
	int ans=power(3,n-1);
	int v=0;
	for(int i=1;i<=n;i++)
		v=(v+s[i]-'a')%3;
	for(int i=0;i<3;i++)
		f[1][i][i]=1;
	for(int i=2;i<=n;i++)
		for(int j=0;j<3;j++)
			for(int k=0;k<3;k++)
				for(int l=0;l<3;l++)
					if(k!=l)
						f[i][(j+l)%3][l]=(f[i][(j+l)%3][l]+f[i-1][j][k])%mod;
	for(int i=0;i<3;i++)
		ans=(ans+mod-f[n][v][i])%mod;
	fl=1;
	for(int i=1;i<n;i++)
		if(s[i]==s[i+1]) 
			fl=0;
	printf("%d\n",(ans+fl)%mod);
	return 0;
}

```