//余事象.. A or B = |A| + |B| - |A and B| (N>=2なのでA=Bか否かで場合分け)
#include <iostream>
#include <atcoder/modint>
using namespace std;
using namespace atcoder;
using mint = modint998244353;

int mod = 998244353;
int n, a, b;
mint fact[200001];

void initFact(int n) {
	fact[0] = 1;
	for (int i = 1; i <= n; i++) {
		fact[i] = i * fact[i - 1];
	}
}

int main() {
	int i;
	cin >> n >> a >> b;
	initFact(n);

	mint ans = fact[n] - 2 * fact[n - 1] + (a != b) * fact[n - 2];
	cout << ans.val() << endl;
	return 0;
}