#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<map>
#include<vector>
#include<ctime>
#include<stack>
#define mp make_pair
#define pa pair<int,int>
#define INF 0x3f3f3f3f
#define inf 0x3f
#define fi first
#define se second
#define pb push_back
#define ll long long
#define ull unsigned long long

using namespace std;

inline ll read()
{
    long long f=1,sum=0;
    char c=getchar();
    while (c<'0' || c>'9')
    {
        if (c=='-') f=-1;
        c=getchar();
    }
    while (c>='0' && c<='9')
    {
        sum=sum*10+c-'0';
        c=getchar();
    }
    return sum*f;
}
const int MAXN=200010;
const int Mod=998244353;
char s[MAXN];
int f[MAXN][3][3][2];
int main()
{
    scanf("%s",s+1);
    int n=strlen(s+1);
    if (n<=3)
    {
        if (n==1)
            cout<<1;
        else if (n==2)
        {
            if (s[1]==s[2])
                cout<<1;
            else
                cout<<2;
        }
        else
        {
            if (s[1]==s[2] || s[2]==s[3])
            {
                if (s[1]==s[2] && s[2]==s[3])
                    cout<<1;
                else 
                    cout<<6;
            }
            else
            {
                if (s[1]==s[3])
                    cout<<7;
                else
                    cout<<3;
            } 
        }
        return 0;
    }
    int sum=0,tot=0;
    for (int i=2;i<=n;i++)
        sum+=(s[i]==s[i-1]),tot+=s[i]-'a';
    tot+=s[1]-'a';
    tot%=3;
    if (sum==n-1)
    {
        cout<<1;
        return 0;
    }
    int ans=!sum,i,j,k,p,last;
    f[1][0][0][0]=f[1][1][1][0]=f[1][2][2][0]=1;
    for(i=2;i<=n;i++)
	    for(j=0;j<3;j++)
		    for(k=0;k<3;k++)
			    for(p=0;p<3;p++)
			    {
				    last=(j-k+3)%3;
				    f[i][j][k][1]=(f[i][j][k][1]+f[i-1][last][p][1])%Mod;
				    if(k==p)	f[i][j][k][1]=(f[i][j][k][1]+f[i-1][last][p][0])%Mod;
				    else	f[i][j][k][0]=(f[i][j][k][0]+f[i-1][last][p][0])%Mod;
			    }
	for(i=0;i<3;i++)	ans=(ans+f[n][tot][i][1])%Mod;
    cout<<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<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<map>
#include<vector>
#include<ctime>
#include<stack>
#define mp make_pair
#define pa pair<int,int>
#define INF 0x3f3f3f3f
#define inf 0x3f
#define fi first
#define se second
#define pb push_back
#define ll long long
#define ull unsigned long long

using namespace std;

inline ll read()
{
    long long f=1,sum=0;
    char c=getchar();
    while (c<'0' || c>'9')
    {
        if (c=='-') f=-1;
        c=getchar();
    }
    while (c>='0' && c<='9')
    {
        sum=sum*10+c-'0';
        c=getchar();
    }
    return sum*f;
}
const int MAXN=200010;
const int Mod=998244353;
char s[MAXN];
int f[MAXN][3][3][2];
int main()
{
    scanf("%s",s+1);
    int n=strlen(s+1);
    if (n<=3)
    {
        if (n==1)
            cout<<1;
        else if (n==2)
        {
            if (s[1]==s[2])
                cout<<1;
            else
                cout<<2;
        }
        else
        {
            if (s[1]==s[2] || s[2]==s[3])
            {
                if (s[1]==s[2] && s[2]==s[3])
                    cout<<1;
                else 
                    cout<<6;
            }
            else
            {
                if (s[1]==s[3])
                    cout<<7;
                else
                    cout<<3;
            } 
        }
        return 0;
    }
    int sum=0,tot=0;
    for (int i=2;i<=n;i++)
        sum+=(s[i]==s[i-1]),tot+=s[i]-'a';
    tot+=s[1]-'a';
    tot%=3;
    if (sum==n-1)
    {
        cout<<1;
        return 0;
    }
    int ans=!sum,i,j,k,p,last;
    f[1][0][0][0]=f[1][1][1][0]=f[1][2][2][0]=1;
    for(i=2;i<=n;i++)
	    for(j=0;j<3;j++)
		    for(k=0;k<3;k++)
			    for(p=0;p<3;p++)
			    {
				    last=(j-k+3)%3;
				    f[i][j][k][1]=(f[i][j][k][1]+f[i-1][last][p][1])%Mod;
				    if(k==p)	f[i][j][k][1]=(f[i][j][k][1]+f[i-1][last][p][0])%Mod;
				    else	f[i][j][k][0]=(f[i][j][k][0]+f[i-1][last][p][0])%Mod;
			    }
	for(i=0;i<3;i++)	ans=(ans+f[n][tot][i][1])%Mod;
    cout<<ans;
    return 0;
}

```