#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define mk make_pair
#define rint register int
#define int long long 
using namespace std;
const int mod = 998244353;
inline int read(){int w=1,s=0;char ch=getchar();while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();}while(isdigit(ch)){s=s*10+ch-'0';ch=getchar();}return w*s;}
int n,f[300010][3][3][2];
char a[1000010];
inline int Not(){
	int fl=1;
	for(rint i=2;i<=n;++i)
	{
		if(a[i]==a[i-1]) fl=0; 
	}return fl;
}
signed main()
{
	scanf("%s",a+1);n=strlen(a+1);
	int fl=1;
	for(rint i=2;i<=n;++i) if(a[i]!=a[i-1]) fl=0;
	if(fl){
		cout<<1;return 0;
	}
	if(n<=3){
	    if(n==2) puts("2");
	    else{
	      if(a[1]!=a[2]&&a[1]!=a[3]&&a[2]!=a[3]) puts("3");
	      else if(a[1]==a[3]) puts("7");
	      else puts("6");
	    }
	    return 0;
    }
	int Sum=0;
	for(rint i=1;i<=n;++i) Sum+=(a[i]-'a');
	Sum%=3;
	f[1][0][0][0]=f[1][1][1][0]=f[1][2][2][0]=1;
	for(rint i=1;i<n;++i)
		for(rint j=0;j<=2;++j)
			for(rint k=0;k<=2;++k)
				for(rint p=0;p<2;++p)
					for(rint S=0;S<=2;++S)
						(f[i+1][(j+S)%3][S][p|(S==k)]+=f[i][j][k][p])%=mod;
	cout<<((f[n][Sum][0][1]+f[n][Sum][1][1])%mod+f[n][Sum][2][1]+Not())%mod;
	return 0;
}
### 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>
#define ll long long
#define pb push_back
#define mk make_pair
#define rint register int
#define int long long 
using namespace std;
const int mod = 998244353;
inline int read(){int w=1,s=0;char ch=getchar();while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();}while(isdigit(ch)){s=s*10+ch-'0';ch=getchar();}return w*s;}
int n,f[300010][3][3][2];
char a[1000010];
inline int Not(){
	int fl=1;
	for(rint i=2;i<=n;++i)
	{
		if(a[i]==a[i-1]) fl=0; 
	}return fl;
}
signed main()
{
	scanf("%s",a+1);n=strlen(a+1);
	int fl=1;
	for(rint i=2;i<=n;++i) if(a[i]!=a[i-1]) fl=0;
	if(fl){
		cout<<1;return 0;
	}
	if(n<=3){
	    if(n==2) puts("2");
	    else{
	      if(a[1]!=a[2]&&a[1]!=a[3]&&a[2]!=a[3]) puts("3");
	      else if(a[1]==a[3]) puts("7");
	      else puts("6");
	    }
	    return 0;
    }
	int Sum=0;
	for(rint i=1;i<=n;++i) Sum+=(a[i]-'a');
	Sum%=3;
	f[1][0][0][0]=f[1][1][1][0]=f[1][2][2][0]=1;
	for(rint i=1;i<n;++i)
		for(rint j=0;j<=2;++j)
			for(rint k=0;k<=2;++k)
				for(rint p=0;p<2;++p)
					for(rint S=0;S<=2;++S)
						(f[i+1][(j+S)%3][S][p|(S==k)]+=f[i][j][k][p])%=mod;
	cout<<((f[n][Sum][0][1]+f[n][Sum][1][1])%mod+f[n][Sum][2][1]+Not())%mod;
	return 0;
}
```