#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <ctime>
#include <cctype>
using namespace std;
const int maxn=1210;
const long long mo=998244353;

int n,ans;
bool bz[maxn];
char s[200010];
long long f[200010][3][3][2];

//__attribute__((optimize("-O3")))
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	ans=1;
	bool pd1=true;
	for (int i=1;i<=n-1;i++)
	{
		if (s[i]==s[i+1]) ans=0;
		if (s[i]!=s[i+1]) pd1=false;
	}
	if (pd1==true)
	{
		printf("1\n");
		return 0;
	}
	if (n==1)
	{
		printf("1\n");return 0;
	}
	else
	if (n==2)
	{
		if (s[1]!=s[2]) printf("2\n");
		else printf("1\n");
		return 0;
	}
	else
	if (n==3)
	{
		if (s[1]==s[2] && s[2]==s[3]) printf("1\n");
		else if ((s[1]==s[2] && s[2]!=s[3]) || (s[2]==s[3] && s[1]!=s[2])) printf("6\n");
		else if (s[1]==s[3] && s[2]!=s[1]) printf("7\n");
		else if (s[1]!=s[2] && s[2]!=s[3] && s[3]!=s[1]) printf("3\n");
		return 0;
	}
	int up=0;
	for (int i=1;i<=n;i++)
	{
		if (s[i]=='b') up++;
		if (s[i]=='c') up+=2;
	}
	up=up%3;
	f[1][0][0][0]=1;
	f[1][1][1][0]=1;
	f[1][2][2][0]=1;
	for (int i=1;i<=n-1;i++)
	{
		for (int a=0;a<=2;a++)
		{
			for (int b=0;b<=2;b++)
			{
				for (int c=0;c<=2;c++)
				{
					int now=0;
					if (b==c)
					{
						now=1;
					}
					for (int hf=0;hf<=1;hf++)
					{
						f[i+1][(a+b)%3][b][max(hf,now)]=(f[i+1][(a+b)%3][b][max(hf,now)]+f[i][a][c][hf])%mo;
					}
				}
			}
		}
	}
	for (int i=0;i<=2;i++)
	{
		ans=(ans+f[n][up][i][1])%mo;
	}
	printf("%lld\n",ans);
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <ctime>
#include <cctype>
using namespace std;
const int maxn=1210;
const long long mo=998244353;

int n,ans;
bool bz[maxn];
char s[200010];
long long f[200010][3][3][2];

//__attribute__((optimize("-O3")))
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	ans=1;
	bool pd1=true;
	for (int i=1;i<=n-1;i++)
	{
		if (s[i]==s[i+1]) ans=0;
		if (s[i]!=s[i+1]) pd1=false;
	}
	if (pd1==true)
	{
		printf("1\n");
		return 0;
	}
	if (n==1)
	{
		printf("1\n");return 0;
	}
	else
	if (n==2)
	{
		if (s[1]!=s[2]) printf("2\n");
		else printf("1\n");
		return 0;
	}
	else
	if (n==3)
	{
		if (s[1]==s[2] && s[2]==s[3]) printf("1\n");
		else if ((s[1]==s[2] && s[2]!=s[3]) || (s[2]==s[3] && s[1]!=s[2])) printf("6\n");
		else if (s[1]==s[3] && s[2]!=s[1]) printf("7\n");
		else if (s[1]!=s[2] && s[2]!=s[3] && s[3]!=s[1]) printf("3\n");
		return 0;
	}
	int up=0;
	for (int i=1;i<=n;i++)
	{
		if (s[i]=='b') up++;
		if (s[i]=='c') up+=2;
	}
	up=up%3;
	f[1][0][0][0]=1;
	f[1][1][1][0]=1;
	f[1][2][2][0]=1;
	for (int i=1;i<=n-1;i++)
	{
		for (int a=0;a<=2;a++)
		{
			for (int b=0;b<=2;b++)
			{
				for (int c=0;c<=2;c++)
				{
					int now=0;
					if (b==c)
					{
						now=1;
					}
					for (int hf=0;hf<=1;hf++)
					{
						f[i+1][(a+b)%3][b][max(hf,now)]=(f[i+1][(a+b)%3][b][max(hf,now)]+f[i][a][c][hf])%mo;
					}
				}
			}
		}
	}
	for (int i=0;i<=2;i++)
	{
		ans=(ans+f[n][up][i][1])%mo;
	}
	printf("%lld\n",ans);
}
```