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

#define ll long long
#define sc(x) scanf("%d",&x)
#define _sc(x) scanf("%lld",&x)
#define pf(x) printf("%d",x);
#define _pf(x) printf("%lld",x);
#define FOR(i,a,b) for(int i = a;i<=b;++i)
#define _FOR(i,a,b) for(int i = a;i<b;++i)
#define pb push_back
#define lb lowber_bound
#define ub upper_bound
#define lson(x) ((x) * 2)
#define rson(x) ((x) * 2 + 1)


const ll mod = 1e9+7;
const ll maxn = 1e6+10;
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;

ll pow2(ll a,ll n){ ll res = 1ll;while(n){if(n&1) res = (res*a)%mod;a = (a*a)%mod;n >>= 1;}return res;}

ll gcd(ll a,ll b){ return b?gcd(b,a%b):a; }

int T;
ll n,m;
int a[maxn];
ll in[maxn];
void init(){
	in[1] = 1;
	for(int i = 2;i <= 1000000;++i){
		in[i] = in[i - 1] + i;
	} 
}
void calc(ll x){
	ll need = (x + 1) * x / 4;
	int cnt = 1 , flag = 1;
	while(need){
		/* 
			二分寻找最大的长度
			need - x + cnt ===> 所需要的 need 数量 - 还剩下(x - cnt)填补交替01形成的纯子矩形 
		*/
		int pos = lower_bound(in + 1, in + 1 + 1000000 , need - (x - cnt)) - (in);
 		for(int i = 1;i<=pos;++i){
			a[cnt++] = flag;
		}
		need -= in[pos];
		flag ^= 1;
	}
}


void solve(){
	cin >> n >> m;
	ll sumN = (n + 1) * n / 2  , sumM = (m + 1) * m / 2;
	if((sumN & 1) && (sumM & 1)){
		cout << "No\n";
		return ;
	}
	cout << "Yes\n";
	int flag = 0;
	if(sumM & 1){
		swap(n,m);
		flag = 1;
	}
	calc(m);
	if(flag){
		for(int i = 1;i <= m;++i){
			for(int j = 1;j<=n;++j){
				cout << a[i] << " \n"[j == n];
			}
		}
	}else{
		for(int i = 1;i <= n;++i){
			for(int j = 1;j<= m;++j){
				cout << a[j] << " \n"[j == m];
			}
		}		
	}

}
int main(void){
	std::ios::sync_with_stdio(0);
	std::cin.tie(0);
	init();
	cin >> T;
	while(T--){
		solve();
	}
	
	return 0;
}