/*ID:PROG:LANG:*/
#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define MOD 998244353
#define MAXN 200005
using namespace std;
int n,fg,S;
LL ans,dp[MAXN][3][3][2];
char s[MAXN];
int rd(){
	int x=0,tp=1;char c=getchar();
	while(c<'0' || c>'9'){if(c=='-')tp=-1;c=getchar();}
	while(c>='0' && c<='9'){x=(x*10+c-'0');c=getchar();}
	return x*tp;
}
int main(){
	// freopen("testdata.in","r",stdin);
	// freopen("1.out","w",stdout);
	scanf("%s",s+1);
	n=strlen(s+1);
	for(int i=2;i<=n;i++)fg|=(s[i]!=s[i-1]);
	if(!fg){printf("1\n");return 0;}
	if(n<=3){
		if(n==2)printf("2\n");
		if(n==3){
			if(s[1]!=s[2] && s[2]!=s[3] && s[1]!=s[3])printf("3\n");
			else if(s[1]==s[3])printf("7\n");
			else printf("6\n");
		}
		return 0;
	}
	dp[1][0][0][0]=dp[1][1][1][0]=dp[1][2][2][0]=1,fg=0;
	//dp[i][j][k][p]表示前i位，前i位的和模3意义下是j，这一位的是k，是否有相邻的两位相同
	for(int i=1;i<n;i++)
		for(int j=0;j<3;j++)
			for(int k=0;k<3;k++)
				for(int p=0;p<2;p++)
					for(int nex=0;nex<3;nex++)
						(dp[i+1][(j+nex)%3][nex][(k==nex)|p]+=dp[i][j][k][p])%=MOD;
	for(int i=1;i<=n;i++)S+=s[i]-'a';S%=3;
	for(int i=0;i<3;i++)(ans+=dp[n][S][i][1])%=MOD;
	for(int i=2;i<=n;i++)fg|=(s[i]==s[i-1]);//S==T
	if(!fg)ans=(ans+1)%MOD;
	printf("%lld\n",ans);
	// printf("%.2lf\n",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}
### 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
/*ID:PROG:LANG:*/
#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define MOD 998244353
#define MAXN 200005
using namespace std;
int n,fg,S;
LL ans,dp[MAXN][3][3][2];
char s[MAXN];
int rd(){
	int x=0,tp=1;char c=getchar();
	while(c<'0' || c>'9'){if(c=='-')tp=-1;c=getchar();}
	while(c>='0' && c<='9'){x=(x*10+c-'0');c=getchar();}
	return x*tp;
}
int main(){
	// freopen("testdata.in","r",stdin);
	// freopen("1.out","w",stdout);
	scanf("%s",s+1);
	n=strlen(s+1);
	for(int i=2;i<=n;i++)fg|=(s[i]!=s[i-1]);
	if(!fg){printf("1\n");return 0;}
	if(n<=3){
		if(n==2)printf("2\n");
		if(n==3){
			if(s[1]!=s[2] && s[2]!=s[3] && s[1]!=s[3])printf("3\n");
			else if(s[1]==s[3])printf("7\n");
			else printf("6\n");
		}
		return 0;
	}
	dp[1][0][0][0]=dp[1][1][1][0]=dp[1][2][2][0]=1,fg=0;
	//dp[i][j][k][p]表示前i位，前i位的和模3意义下是j，这一位的是k，是否有相邻的两位相同
	for(int i=1;i<n;i++)
		for(int j=0;j<3;j++)
			for(int k=0;k<3;k++)
				for(int p=0;p<2;p++)
					for(int nex=0;nex<3;nex++)
						(dp[i+1][(j+nex)%3][nex][(k==nex)|p]+=dp[i][j][k][p])%=MOD;
	for(int i=1;i<=n;i++)S+=s[i]-'a';S%=3;
	for(int i=0;i<3;i++)(ans+=dp[n][S][i][1])%=MOD;
	for(int i=2;i<=n;i++)fg|=(s[i]==s[i-1]);//S==T
	if(!fg)ans=(ans+1)%MOD;
	printf("%lld\n",ans);
	// printf("%.2lf\n",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}
```