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

const int N=2e5+5;
const int MOD=998244353;
int n,t,sm,ans;
char s[N];
int a[N];
int used[3][3][3];
int dp[N][3][3][2];

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

void dfs(int x,int y,int z){
	if(used[x][y][z]) return;
	used[x][y][z]=1;ans++;
	if(x!=y) dfs(3-x-y,3-x-y,z);
	if(y!=z) dfs(x,3-y-z,3-y-z);
}

int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	for(int i=1;i<=n;i++){
		a[i]=s[i]-'a';
		t+=a[i];
	}
	t%=3;
	for(int i=1;i<n;i++) if(a[i]==a[i+1]) sm++;
	if(sm==n-1){
		puts("1");
		return 0;
	}
	if(n==1) puts("1");
	else if(n==2) puts("2");
	else if(n<=3){
		dfs(a[1],a[2],a[3]);
		printf("%d\n",ans);
	}else{
		if(sm==0) ans++;
		for(int i=0;i<3;i++) dp[1][i][i][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 o=0;o<2;o++) for(int l=0;l<3;l++)
			add(dp[i+1][l][(l+k)%3][o|(l==j)],dp[i][j][k][o]);
		for(int i=0;i<3;i++) add(ans,dp[n][i][t][1]);
		printf("%d\n",ans);
	}
	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>
using namespace std;

const int N=2e5+5;
const int MOD=998244353;
int n,t,sm,ans;
char s[N];
int a[N];
int used[3][3][3];
int dp[N][3][3][2];

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

void dfs(int x,int y,int z){
	if(used[x][y][z]) return;
	used[x][y][z]=1;ans++;
	if(x!=y) dfs(3-x-y,3-x-y,z);
	if(y!=z) dfs(x,3-y-z,3-y-z);
}

int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	for(int i=1;i<=n;i++){
		a[i]=s[i]-'a';
		t+=a[i];
	}
	t%=3;
	for(int i=1;i<n;i++) if(a[i]==a[i+1]) sm++;
	if(sm==n-1){
		puts("1");
		return 0;
	}
	if(n==1) puts("1");
	else if(n==2) puts("2");
	else if(n<=3){
		dfs(a[1],a[2],a[3]);
		printf("%d\n",ans);
	}else{
		if(sm==0) ans++;
		for(int i=0;i<3;i++) dp[1][i][i][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 o=0;o<2;o++) for(int l=0;l<3;l++)
			add(dp[i+1][l][(l+k)%3][o|(l==j)],dp[i][j][k][o]);
		for(int i=0;i<3;i++) add(ans,dp[n][i][t][1]);
		printf("%d\n",ans);
	}
	return 0;
}
```