Problem: A mad scientist Dr.Jubal has made a competitive programming task. Try to solve it!

You are given integers n,k. Construct a grid A with size n × n consisting of integers 0 and 1. The very important condition should be satisfied: the sum of all elements in the grid is exactly k. In other words, the number of 1 in the grid is equal to k.

Let's define:

  * A_{i,j} as the integer in the i-th row and the j-th column. 
  * R_i = A_{i,1}+A_{i,2}+...+A_{i,n} (for all 1 ≤ i ≤ n). 
  * C_j = A_{1,j}+A_{2,j}+...+A_{n,j} (for all 1 ≤ j ≤ n). 
  * In other words, R_i are row sums and C_j are column sums of the grid A. 
  * For the grid A let's define the value f(A) = (max(R)-min(R))^2 + (max(C)-min(C))^2 (here for an integer sequence X we define max(X) as the maximum value in X and min(X) as the minimum value in X). 



Find any grid A, which satisfies the following condition. Among such grids find any, for which the value f(A) is the minimum possible. Among such tables, you can find any.

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines contain descriptions of test cases.

For each test case the only line contains two integers n, k (1 ≤ n ≤ 300, 0 ≤ k ≤ n^2).

It is guaranteed that the sum of n^2 for all test cases does not exceed 10^5.

Output

For each test case, firstly print the minimum possible value of f(A) among all tables, for which the condition is satisfied.

After that, print n lines contain n characters each. The j-th character in the i-th line should be equal to A_{i,j}.

If there are multiple answers you can print any.

Example

Input


4
2 2
3 8
1 0
4 16


Output


0
10
01
2
111
111
101
0
0
0
1111
1111
1111
1111

Note

In the first test case, the sum of all elements in the grid is equal to 2, so the condition is satisfied. R_1 = 1, R_2 = 1 and C_1 = 1, C_2 = 1. Then, f(A) = (1-1)^2 + (1-1)^2 = 0, which is the minimum possible value of f(A).

In the second test case, the sum of all elements in the grid is equal to 8, so the condition is satisfied. R_1 = 3, R_2 = 3, R_3 = 2 and C_1 = 3, C_2 = 2, C_3 = 3. Then, f(A) = (3-2)^2 + (3-2)^2 = 2. It can be proven, that it is the minimum possible value of f(A).

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class code6 {

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	public static void main(String[] args) throws CloneNotSupportedException {
		// TODO Auto-generated method stub
		FastReader scn = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		int t = scn.nextInt();
		//int p = t;
		while(t-->0) {
			int n = scn.nextInt();
			int k = scn.nextInt();
			int[][] arr = new int[n][n];
//			int a = 0;
//			int b = 0;
//			int p = 0;
//			while(k>0) {
//				arr[a][b] = 1;
//				a = (a+1)%n;
//				if(b == n-1) {
//					b = (p+1)%n;
//					p++;
//				}else
//					b = (b+1)%n;
//				k--;
//			}
			for(int i=0; i<n; i++) {
				int a = 0;
				int b = i;
				int j = 0;
				while(k>0 && j<n) {
					arr[a][b] = 1;
					a = (a+1)%n;
					b = (b+1)%n;
					k--;
					j++;
				}
				if(k==0)
					break;
			}
			
			int maxx = 0;
			int minx = Integer.MAX_VALUE;
			int maxy = 0;
			int miny = Integer.MAX_VALUE;
			
			for(int i=0; i<n; i++) {
				int c = 0;
				for(int j=0; j<n; j++)
					c += arr[i][j];
				maxx = Math.max(maxx, c);
				minx = Math.min(minx, c);
			}
			for(int i=0; i<n; i++) {
				int c = 0;
				for(int j=0; j<n; j++)
					c += arr[j][i];
				maxy = Math.max(maxy, c);
				miny = Math.min(miny, c);
			}
			
			long ans = 1L*(maxx-minx)*(maxx-minx) + 1L*(maxy-miny)*(maxy-miny);
			out.println(ans);
			for(int i=0; i<n; i++) {
				for(int j=0; j<n; j++)
					out.print(arr[i][j]);
				out.println();
			}
			
			
		}
		
		out.flush();
	}
	
  
	public static boolean check(int mid, int c, int[] arr, int k) {
		int ans = 0;
		for(int i=0; i<arr.length; i++) {
			if(c==0) {
				ans++;
				c = (c+1)%2;
			}else {
				if(arr[i]<=mid) {
					ans++;
					c = (c+1)%2;
				}
			}
		}
		return ans>=k;
	}
	public static int power(int x, int y, int p) 
    { 
        // Initialize result 
        int res = 1;      
         
        // Update x if it is more   
        // than or equal to p 
        x = x % p;  
  
       if (x == 0) return 0; // In case x is divisible by p;    
  
        while (y > 0) 
        { 
            // If y is odd, multiply x 
            // with result 
            if((y & 1)==1) 
                res = (res * x) % p; 
      
            // y must be even now 
            // y = y / 2 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    } 
	
	public static class Pair implements Comparable<Pair>{
		int i;
		int t;
		int a;
		int b;
		Pair(int i,int m, int s, int k){
			this.i = i;
			t = m;
			a = s;
			b = k;
		}
		public int compareTo(Pair p) {
			return this.t-p.t;
		}
	}
    
}