#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
//using mint = modint1000000007;
//const int mod = 1000000007;
using mint = modint998244353;
const int mod = 998244353;
//const int INF = 1e9;
//const long long LINF = 1e18;
//const bool debug = false;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i,l,r)for(int i=(l);i<(r);++i)
#define rrep(i, n) for (int i = (n-1); i >= 0; --i)
#define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i)
#define all(x) (x).begin(),(x).end()
#define allR(x) (x).rbegin(),(x).rend()
#define endl "\n"
#define P pair<int,int>
template<typename A, typename B> inline bool chmax(A & a, const B & b) { if (a < b) { a = b; return true; } return false; }
template<typename A, typename B> inline bool chmin(A & a, const B & b) { if (a > b) { a = b; return true; } return false; }
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		assert(n < mod);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) {
		return com(n, k);
	}
	mint com(int n, int k) {
		//負の二項係数を考慮する場合にコメントアウトを外す
		//if (n < 0) return com(-n, k) * (k % 2 ? -1 : 1);
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
	mint inv(int n, int k) {
		//if (n < 0) return inv(-n, k) * (k % 2 ? -1 : 1);
		if (k < 0 || k > n) return 0;
		return ifact[n] * fact[k] * fact[n - k];
	}
	mint p(int n, int k) {
		return fact[n] * ifact[n - k];
	}
}c(500005);
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, m, k; cin >> n >> m >> k;
	mint ans1 = 0, ans2 = 0, ans3 = 0, ans4 = 0;
	mint inv2 = inv_mod(2, mod);
	{
		//中中
		mint tmp = ((mint)k - 2) * (k - 3);
		ans1 += tmp * (tmp - 1) *inv2 - c(k - 2, 2);
		ans2 += c(k - 2, 2);
	}
	{
		//中右
		mint sub = (k - 2) *(k - 3);
		ans2 += sub * (n - (k - 2)) * 1;
		ans3 += sub * (n - (k - 2)) * (k - 2 - 1);
	}
	{
		//中下
		mint sub = (k - 2) *(k - 3);
		ans2 += sub * (m - (k - 2)) * 1;
		ans3 += sub * (m - (k - 2)) * (k - 2 - 1);
	}
	{
		//右右
		ans3 += (k - 2) *c(n - (k - 2), 2);
		mint tmp = ((mint)n - (k - 2)) *(k - 2);
		ans4 += tmp * (tmp - 1) *inv2 - (k - 2) *c(n - (k - 2), 2);
	}
	{
		//下下
		ans3 += (k - 2) *c(m - (k - 2), 2);
		mint tmp = ((mint)m - (k - 2)) * (k - 2);
		ans4 += tmp * (tmp - 1) *inv2 - (k - 2) *c(m - (k - 2), 2);
	}
	{
		//右下
		// nop
	}
	//cout << ans1.val() << endl;
	//cout << ans2.val() << endl;
	//cout << ans3.val() << endl;
	//cout << ans4.val() << endl;
	mint ans = ans1 + ans2 / 2 + ans3 / 3 + ans4 / 4;
	ans *= c.fact[k - 2];
	ans *= c(n, k - 2);
	ans *= c(m, k - 2);
	cout << ans.val() << endl;
	return 0;
}