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.util.*;		
import java.util.function.Function;
import java.util.stream.Collectors;
import java.io.*;
import java.math.*;
public class Main7{
	static public void main(String args[])throws IOException{
		int tt=i();
		StringBuilder sb=new StringBuilder();
		for(int ttt=1;ttt<=tt;ttt++){
			int n=i();
			int k=i();
			int curr=0;
			int[][] a=new int[n][n];
			for(int diff=0;diff<n;diff++){
				int row=0;
				int col=diff;
				int count=0;
				while(count<n && curr<k){
					if(a[row%n][col%n]==1){
						break;
					}
					a[row%n][col%n]=1;
					row++;
					col++;
					count++;
					curr++;
					if(curr==k){
						break;
					}
				}
				if(curr==k){
					break;
				}
			}
			int[] row=new int[n];
			int[] col=new int[n];
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					if(a[i][j]==1){
						row[i]++;
						col[j]++;
					}
				}
			}
			int maxCol=0,maxRow=0,minCol=Integer.MAX_VALUE,minRow=Integer.MAX_VALUE;
			for(int i=0;i<n;i++){
				maxRow=Math.max(maxRow,row[i]);
				maxCol=Math.max(maxCol,col[i]);
				minRow=Math.min(minRow,row[i]);
				minCol=Math.min(minCol,col[i]);
			}
			long ans=0;
			ans=(maxRow-minRow)*(maxRow-minRow)+(maxCol-minCol)*(maxCol-minCol);
			sb.append(ans+"\n");
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					sb.append(a[i][j]+"");
				}
				sb.append("\n");
			}
		}
		System.out.print(sb.toString());
	}
	static InputReader in=new InputReader(System.in);
	static OutputWriter out=new OutputWriter(System.out);
	static ArrayList<ArrayList<Integer>> graph;
	static int mod=1000000007;
	static class Pair{
		long x;
		int y;
		Pair(long x,int y){
			this.x=x;
			this.y=y;
		}
		/*@Override
		public int hashCode() 
		{ 
			final int temp = 14; 
			int ans = 1; 
			ans =x*31+y*13; 
			return (int)ans;
		}
       @Override
        public boolean equals(Object o) 
        { 
            if (this == o) { 
                return true; 
            } 
            if (o == null) { 
                return false; 
            } 
            if (this.getClass() != o.getClass()) { 
                return false; 
            } 
            Pair other = (Pair)o; 
            if (this.x != other.x || this.y!=other.y) { 
                return false; 
            } 
            return true; 
        }*/
	}
	public static int[] sort(int[] a){
		int n=a.length;
		ArrayList<Integer> ar=new ArrayList<>();
		for(int i=0;i<a.length;i++){
			ar.add(a[i]);
		}
		Collections.sort(ar);
		for(int i=0;i<n;i++){
			a[i]=ar.get(i);
		}
		return a;
	}
	public static long pow(long a, long b){
        long result=1;
        while(b>0){
            if (b % 2 != 0){
                result=(result*a)%mod;
                b--;
            } 
            a=(a*a)%mod;
            b /= 2;
        }   
        return result;
    }
	public static long gcd(long a, long b){
        if (a == 0){
            return b;
		}
        return gcd(b%a, a);
    }
	public static long lcm(long a, long b){
		return a*(b/gcd(a,b));
	}
	public static long l(){
		String s=in.String();
		return Long.parseLong(s);
	}
	public static void pln(String value){
		System.out.println(value);
	}
	public static int i(){
		return in.Int();
	}
	public static String s(){
		return in.String();
	}
}
class InputReader 
{
	private InputStream stream;
	private byte[] buf = new byte[1024];
	private int curChar;
	private int numChars;
	private SpaceCharFilter filter;
	public InputReader(InputStream stream) {
		this.stream = stream;
	}
	public int read() {
		if (numChars== -1)
			throw new InputMismatchException();
		if (curChar >= numChars) {
			curChar = 0;
			try {
				numChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (numChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}
	public int Int() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
	public String String() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isSpaceChar(c));
		return res.toString();
	} 
	public boolean isSpaceChar(int c) {
		if (filter != null)
			return filter.isSpaceChar(c);
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	} 
	public String next() {
		return String();
	} 
	public interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}
}
class OutputWriter {
	private final PrintWriter writer;
	public OutputWriter(OutputStream outputStream) {
		writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
	}
	public OutputWriter(Writer writer) {
		this.writer = new PrintWriter(writer);
	}
	public void print(Object...objects) {
		for (int i = 0; i < objects.length; i++) {
			if (i != 0)
				writer.print(' ');
			writer.print(objects[i]);
		}
	}
	public void printLine(Object...objects) {
		print(objects);
		writer.println();
	}
	public void close() {
		writer.close();
	}
	public void flush() {
		writer.flush();
	}
 }
class IOUtils {
	public static int[] readIntArray(InputReader in, int size) {
		int[] array = new int[size];
		for (int i = 0; i < size; i++)
			array[i] = in.Int();
		return array;
	}
}