#include <bits/stdc++.h>
#define MOD 998244353
using namespace std;

int n,dp[200010][3][3][2];
string s;

inline void add(int &x,int y)
{
	x+=y;
	if(x>=MOD) x-=MOD;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin>>s;
	n=s.size();
	if(n==2){
		if(s[0]==s[1]) cout<<1<<endl;
		else cout<<2<<endl;
		return 0;
	}
	if(n==3){
		if((s[0]==s[1])&&(s[1]==s[2])) cout<<1<<endl;
		else if((s[0]==s[1])&&(s[1]!=s[2])) cout<<6<<endl;
		else if((s[0]==s[2])&&(s[0]!=s[1])) cout<<7<<endl;
		else if((s[1]==s[2])&&(s[0]!=s[2])) cout<<6<<endl;
		else cout<<3<<endl;
		return 0;
	}
	bool flag=false;
	for(int i=0;i<n-1;i++) if(s[i]!=s[i+1]){
		flag=true;
		break;
	}
	if(!flag){
		cout<<1<<endl;
		return 0;
	}
	int ans=1;
	for(int i=0;i<n-1;i++) if(s[i]==s[i+1]){
		ans--;break;
	}
	int num=0;
	for(int i=0;i<n;i++){
		num+=s[i]-'a';
	}
	num%=3;
	dp[0][0][0][0]=1;
	dp[0][1][1][0]=1;
	dp[0][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 l=0;l<3;l++){
					add(dp[i][(k+j)%3][j][1],dp[i-1][k][l][1]);
					if(j==l) add(dp[i][(k+j)%3][j][1],dp[i-1][k][l][0]);
					else add(dp[i][(k+j)%3][j][0],dp[i-1][k][l][0]);
				}
			}
		}
	}
	for(int i=0;i<3;i++) add(ans,dp[n-1][num][i][1]);
	cout<<ans<<endl;
	return 0;
}
### Prompt

Please formulate 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>
#define MOD 998244353
using namespace std;

int n,dp[200010][3][3][2];
string s;

inline void add(int &x,int y)
{
	x+=y;
	if(x>=MOD) x-=MOD;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin>>s;
	n=s.size();
	if(n==2){
		if(s[0]==s[1]) cout<<1<<endl;
		else cout<<2<<endl;
		return 0;
	}
	if(n==3){
		if((s[0]==s[1])&&(s[1]==s[2])) cout<<1<<endl;
		else if((s[0]==s[1])&&(s[1]!=s[2])) cout<<6<<endl;
		else if((s[0]==s[2])&&(s[0]!=s[1])) cout<<7<<endl;
		else if((s[1]==s[2])&&(s[0]!=s[2])) cout<<6<<endl;
		else cout<<3<<endl;
		return 0;
	}
	bool flag=false;
	for(int i=0;i<n-1;i++) if(s[i]!=s[i+1]){
		flag=true;
		break;
	}
	if(!flag){
		cout<<1<<endl;
		return 0;
	}
	int ans=1;
	for(int i=0;i<n-1;i++) if(s[i]==s[i+1]){
		ans--;break;
	}
	int num=0;
	for(int i=0;i<n;i++){
		num+=s[i]-'a';
	}
	num%=3;
	dp[0][0][0][0]=1;
	dp[0][1][1][0]=1;
	dp[0][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 l=0;l<3;l++){
					add(dp[i][(k+j)%3][j][1],dp[i-1][k][l][1]);
					if(j==l) add(dp[i][(k+j)%3][j][1],dp[i-1][k][l][0]);
					else add(dp[i][(k+j)%3][j][0],dp[i-1][k][l][0]);
				}
			}
		}
	}
	for(int i=0;i<3;i++) add(ans,dp[n-1][num][i][1]);
	cout<<ans<<endl;
	return 0;
}
```