#include<bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n,f[2][3][3][2];
int nw=1,pst=0;
char s[2000010];
int main(){
	scanf("%s",s+1);n=strlen(s+1);
	for(int i=0;i<3;i++)f[nw][i][i][0]=1;
	for(int t=2;t<=n;t++){
		swap(nw,pst);
		memset(f[nw],0,sizeof(f[nw]));
		for(int i=0;i<3;i++)for(int j=0;j<3;j++)for(int k=0;k<2;k++){
			for(int cur=0;cur<3;cur++){
				(f[nw][(i+cur)%3][cur][(j==cur)|k]+=f[pst][i][j][k])%=998244353;
			}
		}
	}
	int tot=0,ans=1,flag=0;
	for(int i=1;i<=n;i++){
		tot+=s[i]-'a';tot%=3;
		if(i>1 && s[i]!=s[i-1])flag=1;
		if(i>1 && s[i]==s[i-1])ans=0;
	}
	if(flag==0){puts("1");return 0;}
	if(n==2){puts("2");return 0;}
	if(n==3){
		if(s[1]==s[2])puts("6");
		else {
			if(s[2]==s[3])puts("6");
			else if(s[3]==s[1])puts("7");
			else puts("3");
		}
		return 0;
	}
	//cout << n << " " << tot << endl;
	for(int i=0;i<3;i++)(ans += f[nw][tot][i][1])%=998244353;
	cout << ans << endl;
}

### Prompt

In Cpp, your task is to solve 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 mod = 998244353;
int n,f[2][3][3][2];
int nw=1,pst=0;
char s[2000010];
int main(){
	scanf("%s",s+1);n=strlen(s+1);
	for(int i=0;i<3;i++)f[nw][i][i][0]=1;
	for(int t=2;t<=n;t++){
		swap(nw,pst);
		memset(f[nw],0,sizeof(f[nw]));
		for(int i=0;i<3;i++)for(int j=0;j<3;j++)for(int k=0;k<2;k++){
			for(int cur=0;cur<3;cur++){
				(f[nw][(i+cur)%3][cur][(j==cur)|k]+=f[pst][i][j][k])%=998244353;
			}
		}
	}
	int tot=0,ans=1,flag=0;
	for(int i=1;i<=n;i++){
		tot+=s[i]-'a';tot%=3;
		if(i>1 && s[i]!=s[i-1])flag=1;
		if(i>1 && s[i]==s[i-1])ans=0;
	}
	if(flag==0){puts("1");return 0;}
	if(n==2){puts("2");return 0;}
	if(n==3){
		if(s[1]==s[2])puts("6");
		else {
			if(s[2]==s[3])puts("6");
			else if(s[3]==s[1])puts("7");
			else puts("3");
		}
		return 0;
	}
	//cout << n << " " << tot << endl;
	for(int i=0;i<3;i++)(ans += f[nw][tot][i][1])%=998244353;
	cout << ans << endl;
}

```