/*
对于n>=4,m>=4
猜想答案是4x4的块填充而成，还是比较显然的
最终答案只和右下角n%4，m%4的块的填充情况有关
分类讨论即可
以下是一个合适的4x4块
1101
1110
1011
0111

对于n<=3或m<=3的情况，特判
n=3,n<m
1110
1011
1110
*/
#include<bits/stdc++.h>
using namespace std;
int n, m;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T;
	cin >> T;
	while (T--) {
		cin >> n >> m;
		if (n <= 3 && m <= 3) {
			cout << n * m << "\n";
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < m; j++) {
					cout << 1;
				}
				cout << "\n";
			}
			continue;
		}
		else if (n == 2) {
			if (m % 4 <= 1) cout << (m/4) * 6 + 2 * (m%4) << "\n";
			else cout << (m/4) * 6 + 2 * (m%4) - 1 << "\n";
			for (int i = 0; i < m; i++) cout << (i%4==3?0:1); cout << "\n";
			for (int i = 0; i < m; i++) cout << (i%4==1?0:1); cout << "\n";
			continue;
		}
		else if (n == 3) {
			if (m % 4 <= 1) cout << (m/4) * 9 + 3 * (m%4) << "\n";
			else cout << (m/4) * 9 + 3 * (m%4) - 1 << "\n";
			for (int i = 0; i < m; i++) cout << (i%4==3?0:1); cout << "\n";
			for (int i = 0; i < m; i++) cout << (i%4==1?0:1); cout << "\n";
			for (int i = 0; i < m; i++) cout << (i%4==3?0:1); cout << "\n";
			continue;
		}
		else if (m == 2) {
			if (n % 4 <= 1) cout << (n/4) * 6 + 2 * (n%4) << "\n";
			else cout << (n/4) * 6 + 2 * (n%4) - 1 << "\n";
			for (int i = 0; i < n; i++) {
				cout << (i%4==3?0:1) << (i%4==1?0:1) << "\n";
			}
			continue;
		}
		else if (m == 3) {
			if (n % 4 <= 1) cout << (n/4) * 9 + 3 * (n%4) << "\n";
			else cout << (n/4) * 9 + 3 * (n%4) - 1 << "\n";
			for (int i = 0; i < n; i++) {
				cout << (i%4==3?0:1) << (i%4==1?0:1) << (i%4==3?0:1) << "\n";
			}
			continue;
		}
		
		vector<vector<int>> b(n+4, vector<int>(m));
		vector<int> sum(n+4);
		int co = 0;
		for (int i = 0; i < m; i++) {//构建出第1行
			co += (b[0][i]=(i%4==2?0:1));
		}
		sum[0] = co; co = 0;
		for (int i = 0; i < m; i++) {//构建出第2行
			co += (b[1][i]=(i%4==3?0:1));
		}
		sum[1] = co; co = 0;
		for (int i = 0; i < m; i++) {//构建出第3行
			co += (b[2][i]=(i%4==1?0:1));
		}
		sum[2] = co; co = 0;
		for (int i = 0; i < m; i++) {//构建出第4行
			co += (b[3][i]=(i%4==0?0:1));
		}
		sum[3] = co; co = 0;
		for (int i = 4; i < n+1; i++) {//循环构造至第n+1行
			for (int j = 0; j < m; j++) b[i][j] = b[i-4][j];
			sum[i] = sum[i-4];
		}
		for (int i = 1; i < n+1; i++) sum[i] += sum[i-1];
		
		int pos = 0;//1101为第一行
		if ((m%4==2 && n%4==0) || (m%4==3 && n%4==1)) pos = 1;//1110为第一行
		if (pos == 0) cout << sum[n-1] << "\n";
		else cout << sum[n] - sum[0] << "\n";
		for (int i = pos; i < pos + n; i++) {
			for (int j = 0; j < m; j++) {
				cout << b[i][j];
			}
			cout << "\n";
		}
	}
	return 0;
}
/*
input                                                                             
1
3 7
output
17
1110111
1011101
1110111
*/