#include <bits/stdc++.h>
using namespace std;

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second
#define FILL(a, b) memset(a, b, sizeof(a))

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

const int mod = 998244353;
int add(int a, int b){
	return (a + b < mod) ? (a + b) : (a + b - mod);
}
int mult(int a, int b){
	return 1LL * a * b % mod;
}
int binpow(int a, int n){
	int res = 1;
	while(n){
		if(n & 1)
			res = mult(res, a);
		a = mult(a, a);
		n /= 2;
	}
	return res;
}


const int N = 1000;
int fact[N], ober[N];


map<int, int> dp[N][N];

int F(int n, int m, int k)
{
	if (dp[n][m].count(k)) return dp[n][m][k];
	//if (n == 0 && m == 0) return (k == 0); // ?
	if(n == 0) return (k == 0);
	
	dp[n][m][k] = 0;
	FOR(sz, 0, m + 1){
		int nk = k;
		if(sz % 2){
			if(n % 2)
				nk--;
			else
				nk++;
		}
		
		dp[n][m][k] += F(n - 1, sz, nk);
	}
	return dp[n][m][k];
}



int C(int n, int k){
	if(k < 0 || k > n)
		return 0;
	return mult(fact[n], mult(ober[k], ober[n - k]));
	
}

void prec(){
	fact[0] = 1;
	FOR(i, 1, N)
		fact[i] = mult(fact[i - 1], i);
		
	ober[N - 1] = binpow(fact[N - 1], mod - 2);
	RFOR(i, N, 1)
		ober[i - 1] = mult(ober[i], i);
}

int solve(int n, int m, int k){
	int l = -(min(n, m) / 2);
	int r = ((min(n, m) + 1) / 2);
	
	return mult(C((n + m) / 2, k - l), C((n + m + 1) / 2, r - k));
}

void fin(int val, int n, int m){
	FOR(i, 0, 20){
		FOR(j, 0, 20){
			int n0 = (n + m) / 2;
			int m0 = (n + m + 1) / 2;

			if(mult(C(n0, i), C(m0, j)) == val){
					cout << n0 << " " << i << " " << m0 << " " << j << " " << val << endl;
					return;
			}
		}
	}
	cerr << "bad" << endl;
	cerr << n << " " << m << " " << val << endl;
	exit(0);
	assert(0);
}


int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	prec();
	
	int t;
	cin >> t;
	while(t--){
		int n, m, k;
		cin >> n >> m >> k;
		if(n > m)
			k = (1LL * n * m) % 2 - k;
			
		cout << solve(n, m, k) << "\n";
	}
	/*
	int n, m;
	cin >> n >> m;
	
	int mn = 10, mx = -10;
	FOR (k, -5, 6)
	{
		F(n, m, k);
		if(dp[n][m][k] != 0){
			mn = min(mn, k);
			mx = max(mx, k);
		}
		cout << k << ' ' << dp[n][m][k] << " " << solve(n, m, k) << '\n';
		assert(dp[n][m][k] == solve(n, m, k));
		fin(dp[n][m][k], n, m);
	}
	cerr << mn << " " << mx << endl;
	*/
	
	return 0;
}