#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5,mod=998244353;
char s[N];
int n;
ll dp[N][3][3][2];
int main()
{
    scanf("%s",s+1);
    n=strlen(s+1);
    bool flag=false;
    for(int i=2;i<=n;i++)
        flag|=s[i]!=s[i-1];
    if(!flag){printf("1\n");return 0;}
    if(n<=3)
    {
        if(n==2) puts("2");
        if(n==3)
        {
            if(s[1]!=s[2]&&s[2]!=s[3]&&s[1]!=s[3]) puts("3");
            else if(s[1]==s[3]) puts("7");
            else puts("6");
        }
        return 0;
    }
    dp[1][0][0][0]=dp[1][1][1][0]=dp[1][2][2][0]=1;
    int sum=0;
    for(int i=1;i<n;i++)
        for(int j=0;j<=2;j++)
            for(int k=0;k<=2;k++)
            for(int h=0;h<2;h++)
            for(int f=0;f<=2;f++)
                (dp[i+1][(j+f)%3][f][(k==f)|h]+=dp[i][j][k][h])%=mod;
    for(int i=1;i<=n;i++) (sum+=s[i]-'a')%=3;
    ll ans=0;
    for(int i=0;i<=2;i++)
        (ans+=dp[n][sum][i][1])%=mod;
    flag=false;
    for(int i=2;i<=n;i++)
        flag|=s[i]==s[i-1];
    if(!flag) ans++;
    printf("%lld\n",ans);
}

### Prompt

Your task is to create 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;
typedef long long ll;
const int N=2e5+5,mod=998244353;
char s[N];
int n;
ll dp[N][3][3][2];
int main()
{
    scanf("%s",s+1);
    n=strlen(s+1);
    bool flag=false;
    for(int i=2;i<=n;i++)
        flag|=s[i]!=s[i-1];
    if(!flag){printf("1\n");return 0;}
    if(n<=3)
    {
        if(n==2) puts("2");
        if(n==3)
        {
            if(s[1]!=s[2]&&s[2]!=s[3]&&s[1]!=s[3]) puts("3");
            else if(s[1]==s[3]) puts("7");
            else puts("6");
        }
        return 0;
    }
    dp[1][0][0][0]=dp[1][1][1][0]=dp[1][2][2][0]=1;
    int sum=0;
    for(int i=1;i<n;i++)
        for(int j=0;j<=2;j++)
            for(int k=0;k<=2;k++)
            for(int h=0;h<2;h++)
            for(int f=0;f<=2;f++)
                (dp[i+1][(j+f)%3][f][(k==f)|h]+=dp[i][j][k][h])%=mod;
    for(int i=1;i<=n;i++) (sum+=s[i]-'a')%=3;
    ll ans=0;
    for(int i=0;i<=2;i++)
        (ans+=dp[n][sum][i][1])%=mod;
    flag=false;
    for(int i=2;i<=n;i++)
        flag|=s[i]==s[i-1];
    if(!flag) ans++;
    printf("%lld\n",ans);
}

```