#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int N=2e5+10,mo=998244353;
int n,sum,tot,ans,f[N][3][3][2];
char s[N];
int read()
{
    int f=1,x=0; char c=getchar();
    while (c<'0'||c>'9'){ if (c=='-') f=-1; c=getchar(); }
    while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
    return x*f;
}
int main()
{
    scanf("%s",s+1),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;
    }
    for (int i=2;i<=n;i++) sum+=(s[i]==s[i-1]),tot+=s[i]-'a';
    (tot+=s[1]-'a')%=3,ans=!sum;
    if (sum==n-1){ cout<<1; return 0;}
    f[1][1][1][0]=f[1][2][2][0]=f[1][0][0][0]=1;
    for (int i=2;i<=n;i++) for (int j=0;j<=2;j++) for (int k=0;k<=2;k++) for (int p=0;p<=2;p++)
    {
    	int last=(j-k+3)%3;
        (f[i][j][k][1]+=f[i-1][last][p][1])%=mo;
        if (k==p) (f[i][j][k][1]+=f[i-1][last][p][0])%=mo; else (f[i][j][k][0]+=f[i-1][last][p][0])%=mo;
    }
    for (int i=0;i<=2;i++) (ans+=f[n][tot][i][1])%=mo;
    printf("%d\n",ans);
}
### Prompt

Generate 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 <cstring>
#include <iostream>
using namespace std;
const int N=2e5+10,mo=998244353;
int n,sum,tot,ans,f[N][3][3][2];
char s[N];
int read()
{
    int f=1,x=0; char c=getchar();
    while (c<'0'||c>'9'){ if (c=='-') f=-1; c=getchar(); }
    while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
    return x*f;
}
int main()
{
    scanf("%s",s+1),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;
    }
    for (int i=2;i<=n;i++) sum+=(s[i]==s[i-1]),tot+=s[i]-'a';
    (tot+=s[1]-'a')%=3,ans=!sum;
    if (sum==n-1){ cout<<1; return 0;}
    f[1][1][1][0]=f[1][2][2][0]=f[1][0][0][0]=1;
    for (int i=2;i<=n;i++) for (int j=0;j<=2;j++) for (int k=0;k<=2;k++) for (int p=0;p<=2;p++)
    {
    	int last=(j-k+3)%3;
        (f[i][j][k][1]+=f[i-1][last][p][1])%=mo;
        if (k==p) (f[i][j][k][1]+=f[i-1][last][p][0])%=mo; else (f[i][j][k][0]+=f[i-1][last][p][0])%=mo;
    }
    for (int i=0;i<=2;i++) (ans+=f[n][tot][i][1])%=mo;
    printf("%d\n",ans);
}
```