#include<bits/stdc++.h>
#define LL long long 
#define clr(x,i) memset(x,i,sizeof(x))
using namespace std;
const int N=200005,mo=998244353;
char s[N];
int n,f[N][3][3][2],val;//i th, cur j, mod k, same l
inline void add(int &x,int y)
{x+=y; if(x>mo) x-=mo;}
bool chks()
{
	bool ret=1;
	for(int i=1;i<=n;i++)
	{
		val=(val+s[i]-'a')%3;
		if(s[i]!=s[1]) ret=0;
	}
	return ret;
}
int main()
{
	scanf("%s",s+1); n=strlen(s+1);
	if(chks()){printf("1\n");return 0;}
	if(n==2){printf("2\n");return 0;}
	if(n==3){
//		aab - acc - bbc - baa - cca - cbb | aab
//		abc - ccc
//		    - aaa
//		aba - cca ~ bba (+6)
//		    - acc (+0)
		if(s[1]!=s[2]&&s[2]!=s[3]&&s[3]!=s[1]) printf("3\n");
		else if(s[1]==s[3]) printf("7\n");
		else printf("6\n");
		return 0;
	}
	f[1][0][0][0]=f[1][1][1][0]=f[1][2][2][0]=1;
	for(int i=1;i<n;i++){
		for(int j=0;j<3;j++){
			for(int k=0;k<3;k++){
				for(int x=0;x<3;x++){
					int nxtmod=(k+x)%3;
					if(x!=j) add(f[i+1][x][nxtmod][0],f[i][j][k][0]);
					add(f[i+1][x][nxtmod][1],f[i][j][k][1]);
					if(x==j) add(f[i+1][x][nxtmod][1],f[i][j][k][0]);
				}
			}
		}
	}
	int ans=0;
	for(int i=0;i<3;i++)
	  add(ans,f[n][i][val][1]);
	int kd=1;
	for(int i=2;i<=n;i++)
	  if(s[i]==s[i-1]) kd=0;
	add(ans,kd);
	printf("%d\n",ans);
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define LL long long 
#define clr(x,i) memset(x,i,sizeof(x))
using namespace std;
const int N=200005,mo=998244353;
char s[N];
int n,f[N][3][3][2],val;//i th, cur j, mod k, same l
inline void add(int &x,int y)
{x+=y; if(x>mo) x-=mo;}
bool chks()
{
	bool ret=1;
	for(int i=1;i<=n;i++)
	{
		val=(val+s[i]-'a')%3;
		if(s[i]!=s[1]) ret=0;
	}
	return ret;
}
int main()
{
	scanf("%s",s+1); n=strlen(s+1);
	if(chks()){printf("1\n");return 0;}
	if(n==2){printf("2\n");return 0;}
	if(n==3){
//		aab - acc - bbc - baa - cca - cbb | aab
//		abc - ccc
//		    - aaa
//		aba - cca ~ bba (+6)
//		    - acc (+0)
		if(s[1]!=s[2]&&s[2]!=s[3]&&s[3]!=s[1]) printf("3\n");
		else if(s[1]==s[3]) printf("7\n");
		else printf("6\n");
		return 0;
	}
	f[1][0][0][0]=f[1][1][1][0]=f[1][2][2][0]=1;
	for(int i=1;i<n;i++){
		for(int j=0;j<3;j++){
			for(int k=0;k<3;k++){
				for(int x=0;x<3;x++){
					int nxtmod=(k+x)%3;
					if(x!=j) add(f[i+1][x][nxtmod][0],f[i][j][k][0]);
					add(f[i+1][x][nxtmod][1],f[i][j][k][1]);
					if(x==j) add(f[i+1][x][nxtmod][1],f[i][j][k][0]);
				}
			}
		}
	}
	int ans=0;
	for(int i=0;i<3;i++)
	  add(ans,f[n][i][val][1]);
	int kd=1;
	for(int i=2;i<=n;i++)
	  if(s[i]==s[i-1]) kd=0;
	add(ans,kd);
	printf("%d\n",ans);
	return 0;
}

```