#include <iostream>
#include <cstdio>
#define mod 998244353
using namespace std;
string s;
int f[1000006][3][2][5];
void upd(int& x, int y){
	x += y;
	if (x >= mod)
		x -= mod;
}
int main(){
	cin >> s;
	int n = s.length();
	int sum = 0;
	if (n <= 3){
		if (n == 1)
			cout << 1 << endl;
		if (n == 2){
			if (s[0] == s[1])
				cout << 1 << endl;
			else
				cout << 2 << endl;
		}
		if (n == 3){
            if (s[0] == s[1] && s[1] == s[2])
				cout << 1 << endl;
			else 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;
	}
	bool same = false;
	for (int i = 1; i < n; ++ i)
		if (s[i] != s[i - 1]){
			same = true;
			break;
		}
	if (! same){
		cout << 1 << endl;
		return 0;
	}
	for (int i = 0; i < n; ++ i)
		sum = (sum + s[i] - 'a' + 1) % 3;
	f[0][0][0][3] = 1;
	for (int i = 0; i < n; ++ i)
		for (int j = 0; j <= 2; ++ j)
			for (int have = 0; have <= 1; ++ have)
				for (int last = 0; last <= 3; ++ last){
					if (i)
						upd(f[i + 1][(j + last) % 3][1][last], f[i][j][have][last]);
					for (int k = 0; k <= 2; ++ k)
						if (k != last)
							upd(f[i + 1][(j + k) % 3][have][k], f[i][j][have][last]);
				}
	int ans = 0;
	for (int last = 0; last <= 2; ++ last)
		upd(ans, f[n][sum][1][last]);
	bool flag = false;
	for (int i = 1; i < n; ++ i)
		if (s[i] == s[i - 1]){
			flag = true;
			break;
		}
	upd(ans, ! flag);
	cout << ans << endl;
}

### Prompt

Please formulate 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 <iostream>
#include <cstdio>
#define mod 998244353
using namespace std;
string s;
int f[1000006][3][2][5];
void upd(int& x, int y){
	x += y;
	if (x >= mod)
		x -= mod;
}
int main(){
	cin >> s;
	int n = s.length();
	int sum = 0;
	if (n <= 3){
		if (n == 1)
			cout << 1 << endl;
		if (n == 2){
			if (s[0] == s[1])
				cout << 1 << endl;
			else
				cout << 2 << endl;
		}
		if (n == 3){
            if (s[0] == s[1] && s[1] == s[2])
				cout << 1 << endl;
			else 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;
	}
	bool same = false;
	for (int i = 1; i < n; ++ i)
		if (s[i] != s[i - 1]){
			same = true;
			break;
		}
	if (! same){
		cout << 1 << endl;
		return 0;
	}
	for (int i = 0; i < n; ++ i)
		sum = (sum + s[i] - 'a' + 1) % 3;
	f[0][0][0][3] = 1;
	for (int i = 0; i < n; ++ i)
		for (int j = 0; j <= 2; ++ j)
			for (int have = 0; have <= 1; ++ have)
				for (int last = 0; last <= 3; ++ last){
					if (i)
						upd(f[i + 1][(j + last) % 3][1][last], f[i][j][have][last]);
					for (int k = 0; k <= 2; ++ k)
						if (k != last)
							upd(f[i + 1][(j + k) % 3][have][k], f[i][j][have][last]);
				}
	int ans = 0;
	for (int last = 0; last <= 2; ++ last)
		upd(ans, f[n][sum][1][last]);
	bool flag = false;
	for (int i = 1; i < n; ++ i)
		if (s[i] == s[i - 1]){
			flag = true;
			break;
		}
	upd(ans, ! flag);
	cout << ans << endl;
}

```