#include<bits/stdc++.h>
using namespace std;
const int M=2e5+5,mod=998244353;
char ch[M];
int n,sum,i,j,k,a,b,dp[M][3][3][2];
bool flag=1;
void spj()
{
    for(i=2;i<=n;++i)if(ch[i]!=ch[i-1]){flag=0;break;}
    if(flag)puts("1"),exit(0);
    if(n<=3)(n==2?puts("2"):((ch[1]!=ch[2]&&ch[1]!=ch[3]&&ch[2]!=ch[3])?puts("3"):((ch[1]==ch[3])?puts("7"):puts("6")))),exit(0);
}
bool check(){for(int i=1;i<=n;++i)if(ch[i]==ch[i-1])return 0;return 1;}
void in(){scanf("%s",ch+1);}
void ac()
{
    n=strlen(ch+1);spj();
    for(i=1;i<=n;++i)sum=(sum+ch[i]-'a')%3;
    for(i=0;i<=2;++i)dp[1][i][i][0]=1;
    for(i=1;i<=n-1;++i)for(j=0;j<=2;++j)for(k=0;k<=2;++k)for(a=0;a<=1;++a)for(b=0;b<=2;++b)
    (dp[i+1][(j+b)%3][b][a|(k==b)]+=dp[i][j][k][a])%=mod;
    printf("%d",(0ll+dp[n][sum][0][1]+dp[n][sum][1][1]+dp[n][sum][2][1]+check())%mod);
}
int main(){in();ac();}
### 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
#include<bits/stdc++.h>
using namespace std;
const int M=2e5+5,mod=998244353;
char ch[M];
int n,sum,i,j,k,a,b,dp[M][3][3][2];
bool flag=1;
void spj()
{
    for(i=2;i<=n;++i)if(ch[i]!=ch[i-1]){flag=0;break;}
    if(flag)puts("1"),exit(0);
    if(n<=3)(n==2?puts("2"):((ch[1]!=ch[2]&&ch[1]!=ch[3]&&ch[2]!=ch[3])?puts("3"):((ch[1]==ch[3])?puts("7"):puts("6")))),exit(0);
}
bool check(){for(int i=1;i<=n;++i)if(ch[i]==ch[i-1])return 0;return 1;}
void in(){scanf("%s",ch+1);}
void ac()
{
    n=strlen(ch+1);spj();
    for(i=1;i<=n;++i)sum=(sum+ch[i]-'a')%3;
    for(i=0;i<=2;++i)dp[1][i][i][0]=1;
    for(i=1;i<=n-1;++i)for(j=0;j<=2;++j)for(k=0;k<=2;++k)for(a=0;a<=1;++a)for(b=0;b<=2;++b)
    (dp[i+1][(j+b)%3][b][a|(k==b)]+=dp[i][j][k][a])%=mod;
    printf("%d",(0ll+dp[n][sum][0][1]+dp[n][sum][1][1]+dp[n][sum][2][1]+check())%mod);
}
int main(){in();ac();}
```