#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define ll long long
#define double long double
#define lc u<<1
#define rc u<<1|1
#define X first
#define Y second
//#define int long long
const int N = 5e5 + 50;
const int M = 2005;
const ll maxm = 1e18 + 5;
const ll mod = 998244353;




void solve()
{
	///首先，ans是可以由一行递推出来的
	//问题就变成了，对于n/m在一行中是否能找到结论
	//如果是奇数，显然是找不到答案的，如果是偶数
	//我们直接贪心的放，
	ll n, m; cin >> n >> m;
	if ((n) * (n + 1) % 4==0) {
		//cout << "Yes" << endl;
		vector<int>v;
		v.push_back(114514);
		ll sum = (n) * (n + 1) / 4;
		//放纯的(1)如果放不了，就放0
		ll idx = n;
		int ans = 1;
		while (1) {
			if (idx <= 0)break;
			ll val = (idx) * (idx + 1) / 2;
			if (val > sum||val==sum&&idx+v.size()!=n+1) {
				idx--; continue;
			}
			sum -= val;
			for (int i = 1; i <= idx; i++) {
				v.push_back(ans);
			}
			ans = ~ans;
		}
		if (v.size() == n+1) {
			cout << "Yes" << endl;
			for (int i = 1; i <= n; i++) {
				cout << v[i];
				for (int j = 2; j <= m; j++) {
					cout << " " << v[i] << endl;
				}
				cout << endl;
			}
			return;
		}
	}
	if ((m) * (m + 1) % 4==0) {
		//cout << "Yes" << endl;
		vector<int>v;
		v.push_back(114514);
		ll sum = (m) * (m + 1) / 4;
		//放纯的(1)如果放不了，就放0
		ll idx = m;
		int ans = 1;
		while (1) {
			if (idx <= 0)break;
			ll val = (idx) * (idx + 1) / 2;
			if (val > sum||val==sum&&v.size()+idx!=m+1) {
				idx--; continue;
			}
			sum -= val;
			for (int i = 1; i <= idx; i++) {
				v.push_back(ans);
			}
			ans = !ans;
		}
		if (v.size() == m+1) {
			cout << "Yes" << endl;
			for (int i = 1; i <= n; i++) {
				cout << v[1];
				for (int j = 2; j <= m; j++) {
					cout << " " << v[j];
				}
				cout << endl;
			}
			return;
		}
	}
	cout << "No" << endl;
	return;
}


signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int T = 1;
	std::cin >> T;
	while (T--)
		solve();
	return 0;
}


ll ksm(ll a, ll b) {
	ll base = a;
	ll ans = 1;
	while (b) {
		if (b & 1)ans *= base;
		base *= base;
		b >>= 1;
	}
	return ans;
}
ll gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}