#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
#define MAX 201010
#define MAXS 20
#define INF 1000000000000000001
#define bb ' '
#define ln '\n'
#define Ln '\n'
#define X 20
bool ispos(int N, int M, int K) {
if (1ll * N * M > 200000) return false;
if (N & 1) return false;
if (M & 1) return false;
N >>= 1;
M >>= 1;
if (N > M) swap(N, M);
if (K == M + 1 && N == M) return false;
if (K == N * M - 1) return false;
return M <= K && K <= N * M;
}
vector<vector<int>> ans;
void solve(int N, int M, int K, int xpos, int ypos) {
if (!K) return;
if (M == 1 || N == 1) {
if (M == 1) { for (int i = 0; i < N; i++) for (auto x : { i * 2, i * 2 + 1 }) for (auto y : { 0, 1 }) ans[xpos + x][ypos + y] = i + 1; }
else { for (int i = 0; i < M; i++) for (auto x : { 0, 1 }) for (auto y : { 2 * i, 2 * i + 1 }) ans[xpos + x][ypos + y] = i + 1; }
return;
}
if (ispos(N * 2 - 2, M * 2 - 2, K - 1)) {
int i;
for (i = 0; i < N * 2; i++) ans[xpos + i][ypos] = ans[xpos + i][ypos + M * 2 - 1] = K;
for (i = 0; i < M * 2; i++) ans[xpos][ypos + i] = ans[xpos + N * 2 - 1][ypos + i] = K;
solve(N - 1, M - 1, K - 1, xpos + 1, ypos + 1);
return;
}
if (K >= M && ispos(N * 2 - 2, M * 2, K - M)) {
for (int i = 0; i < M; i++) for (auto x : { 0, 1 }) for (auto y : { 2 * i, 2 * i + 1 }) ans[xpos + x][ypos + y] = K - i;
solve(N - 1, M, K - M, xpos + 2, ypos);
return;
}
if (K >= N && ispos(N * 2, M * 2 - 2, K - N)) {
for (int i = 0; i < N; i++) for (auto x : { i * 2, i * 2 + 1 }) for (auto y : { 0, 1 }) ans[xpos + x][ypos + y] = K - i;
solve(N, M - 1, K - N, xpos, ypos + 2);
}
}
signed main() {
ios::sync_with_stdio(false), cin.tie(0);
int T;
cin >> T;
while (T--) {
int N, M;
ll K;
cin >> N >> M >> K;
if (!ispos(N, M, K)) {
cout << "NO" << ln;
continue;
}
ans = vector<vector<int>>(N, vector<int>(M));
N >>= 1;
M >>= 1;
solve(N, M, K, 0, 0);
cout << "YES" << ln;
int i, j;
for (i = 0; i < N * 2; i++) {
for (j = 0; j < M * 2; j++) cout << ans[i][j] << bb;
cout << ln;
}
}
}
