#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=998244353;

int cnt[3];
// dp[index][連続文字の有無][mod][前の文字]
ll dp[202020][2][3][3];

int main(){
  string S; cin>>S;
  int n=S.length(), flag=1;
  int sum=0;
  rep(i,n){
    if(i<n-1) if(S[i+1]==S[i]) flag=0;
    cnt[S[i]-'a']++;
    sum+=(S[i]-'a');
  }
  sum%=3;
  rep(i,3) if(cnt[i]==n){
    cout<<1<<endl;
    return 0;
  }
  if(n==2){
    cout<<2<<endl;
    return 0;
  }
  if(n==3){
    if(S[0]!=S[1]&&S[0]!=S[2]&&S[1]!=S[2]){
      cout<<3<<endl;
    }else if(S[0]==S[2]){
      cout<<7<<endl;
    }else{
      cout<<6<<endl;
    }
    return 0;
  }

  dp[1][0][0][0]=1;
  dp[1][0][1][1]=1;
  dp[1][0][2][2]=1;
  for(int i=1;i<n;i++){
    // j:連続文字, k:mod, l:前の文字
    rep(j,2) rep(k,3) rep(l,3) rep(m,3){
      (dp[i+1][j|l==m][(k+m)%3][m]+=dp[i][j][k][l])%=MOD;
    }
  }
  ll ans=flag;
  rep(i,3) (ans+=dp[n][1][sum][i])%=MOD;
  cout<<ans<<endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=998244353;

int cnt[3];
// dp[index][連続文字の有無][mod][前の文字]
ll dp[202020][2][3][3];

int main(){
  string S; cin>>S;
  int n=S.length(), flag=1;
  int sum=0;
  rep(i,n){
    if(i<n-1) if(S[i+1]==S[i]) flag=0;
    cnt[S[i]-'a']++;
    sum+=(S[i]-'a');
  }
  sum%=3;
  rep(i,3) if(cnt[i]==n){
    cout<<1<<endl;
    return 0;
  }
  if(n==2){
    cout<<2<<endl;
    return 0;
  }
  if(n==3){
    if(S[0]!=S[1]&&S[0]!=S[2]&&S[1]!=S[2]){
      cout<<3<<endl;
    }else if(S[0]==S[2]){
      cout<<7<<endl;
    }else{
      cout<<6<<endl;
    }
    return 0;
  }

  dp[1][0][0][0]=1;
  dp[1][0][1][1]=1;
  dp[1][0][2][2]=1;
  for(int i=1;i<n;i++){
    // j:連続文字, k:mod, l:前の文字
    rep(j,2) rep(k,3) rep(l,3) rep(m,3){
      (dp[i+1][j|l==m][(k+m)%3][m]+=dp[i][j][k][l])%=MOD;
    }
  }
  ll ans=flag;
  rep(i,3) (ans+=dp[n][1][sum][i])%=MOD;
  cout<<ans<<endl;
  return 0;
}

```