#include <bits/stdc++.h>
using namespace std;
#define mod 998244353
long long dp[200010][2][3][3];

int main() {
    string s; cin >> s;
    string t = s;
    int n = s.size();
    sort(t.begin(), t.end());
    if (t.front() == t.back()) {
        cout << 1 << endl;
        return 0;
    }
    if (n == 2) {
        cout << 2 << endl;
        return 0;
    } else if (n == 3) {
        if (s[0] == s[1] || s[1] == s[2]) {
            cout << 6 << endl;
        } else if (s[0] == s[2]) {
            cout << 7 << endl;
        } else {
            cout << 3 << endl;
        }
        return 0;
    }
    int sum = 0;
    bool flag = true;
    for (int i = 0; i < n; i++) {
        sum = (sum + s[i] - 'a') % 3;
        if (i && s[i-1] == s[i]) flag = false; 
    }
    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++) {
        for (int j = 0; j < 2; j++) {
            for (int s = 0; s < 3; s++) {
                for (int prev = 0; prev < 3; prev++) {
                    for (int cur = 0; cur < 3; cur++) {
                        if (j == 0) {
                            if (prev == cur) {
                                dp[i+1][j+1][(s+cur)%3][cur] += dp[i][j][s][prev];
                                dp[i+1][j+1][(s+cur)%3][cur] %= mod;
                            } else {
                                dp[i+1][j][(s+cur)%3][cur] += dp[i][j][s][prev];
                                dp[i+1][j][(s+cur)%3][cur] %= mod;
                            }
                        } else {
                            dp[i+1][j][(s+cur)%3][cur] += dp[i][j][s][prev];
                            dp[i+1][j][(s+cur)%3][cur] %= mod;
                        }
                    }
                }
            }
        }
    }
    cout << (flag+dp[n][1][sum][0]+dp[n][1][sum][1]+dp[n][1][sum][2]) % mod << endl;
    return 0;
}
### 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;
#define mod 998244353
long long dp[200010][2][3][3];

int main() {
    string s; cin >> s;
    string t = s;
    int n = s.size();
    sort(t.begin(), t.end());
    if (t.front() == t.back()) {
        cout << 1 << endl;
        return 0;
    }
    if (n == 2) {
        cout << 2 << endl;
        return 0;
    } else if (n == 3) {
        if (s[0] == s[1] || s[1] == s[2]) {
            cout << 6 << endl;
        } else if (s[0] == s[2]) {
            cout << 7 << endl;
        } else {
            cout << 3 << endl;
        }
        return 0;
    }
    int sum = 0;
    bool flag = true;
    for (int i = 0; i < n; i++) {
        sum = (sum + s[i] - 'a') % 3;
        if (i && s[i-1] == s[i]) flag = false; 
    }
    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++) {
        for (int j = 0; j < 2; j++) {
            for (int s = 0; s < 3; s++) {
                for (int prev = 0; prev < 3; prev++) {
                    for (int cur = 0; cur < 3; cur++) {
                        if (j == 0) {
                            if (prev == cur) {
                                dp[i+1][j+1][(s+cur)%3][cur] += dp[i][j][s][prev];
                                dp[i+1][j+1][(s+cur)%3][cur] %= mod;
                            } else {
                                dp[i+1][j][(s+cur)%3][cur] += dp[i][j][s][prev];
                                dp[i+1][j][(s+cur)%3][cur] %= mod;
                            }
                        } else {
                            dp[i+1][j][(s+cur)%3][cur] += dp[i][j][s][prev];
                            dp[i+1][j][(s+cur)%3][cur] %= mod;
                        }
                    }
                }
            }
        }
    }
    cout << (flag+dp[n][1][sum][0]+dp[n][1][sum][1]+dp[n][1][sum][2]) % mod << endl;
    return 0;
}
```