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.*;import java.util.*;import java.math.*;
public class Main
{
	static long mod=1000000007l;
    static int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;
    static long maxl=Long.MAX_VALUE,minl=Long.MIN_VALUE;
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st;
    
    
    static public void main(String[] args)throws Exception
    {	
    	st=new StringTokenizer(br.readLine());
    	int t=i();
    	StringBuilder sb=new StringBuilder();
    	o:
    	while(t-->0)
    	{
//    		long a=l();
//    		long b=l();
//    		long n=l();
//    		long m=l();
//    		if(min(a,b)>=n&&min(a,b)>=m)pl("Yes");
//    		else pl("No");
    		int n=i();
    		int k=i();
    		int val=n-(k%n);
    		int a=k/n;
    		int b=a+1;
    		int ar[][]=new int[n][n];
    		int i=0;
    		for(int x=0;x<val;x++)
    		{
    			
    			for(int y=1;y<=a;y++)
    			{
    				ar[x][i]=1;
    				i=(i+1)%n;
    			}
    		}
    		for(int x=val;x<n;x++)
    		{
    			
    			for(int y=1;y<=b;y++)
    			{
    				ar[x][i]=1;
    				i=(i+1)%n;
    			}
    		}
    		int mi=max;
    		int mx=min;
    		for(int x=0;x<n;x++)
    		{
    			int c1=0;
    			for(int y=0;y<n;y++)
    			{
    				if(ar[x][y]==1)c1++;
    				//sb.append(ar[x][y]);
    				//sb.append(' ');
    			}
    			mi=min(mi,c1);
    			mx=max(mi,c1);
    			//sb.append("\n");
    		}
    		
    		int mi2=max;
    		int mx2=min;
    		for(int x=0;x<n;x++)
    		{
    			int c1=0;
    			for(int y=0;y<n;y++)
    			{
    				if(ar[y][x]==1)c1++;
    				//sb.append(ar[x][y]);
    				//sb.append(' ');
    			}
    			mi2=min(mi2,c1);
    			mx2=max(mx2,c1);
    		//	sb.append("\n");
    		}
    	//	pl(mx+" "+mi+" "+mx2+" "+mi2);
    		sb.append(((mx-mi)*(mx-mi)+(mx2-mi2)*(mx2-mi2)));
    		sb.append("\n");
    		for(int x=0;x<n;x++)
    		{
    			//int c1=0;
    			for(int y=0;y<n;y++)
    			{
    				//if(ar[y][x]==1)c1++;
    				sb.append(ar[x][y]);
    				//sb.append(' ');
    			}
    			//mi2=min(mi2,c1);
    			//mx2=max(mi2,c1);
    			sb.append("\n");
    		}
    		//sb.append("\n");
    	}
    	p(sb);
    }
    
    static int max(int a,int b){return a>b?a:b;}
    static int min(int a,int b){return a<b?a:b;}
    static int abs(int a){return Math.abs(a);}
    static long max(long a,long b){return a>b?a:b;}
    static long min(long a,long b){return a<b?a:b;}
    static long abs(long a){return Math.abs(a);}
    static int sq(int a){return (int)Math.sqrt(a);}
    static long sq(long a){return (long)Math.sqrt(a);}
    static int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
//    static void g(int i)
//    {
//    	for(int e:ar[i])
//    	{
//    		if(e==i)continue;
//    		al[i].add(e);
//    		g(e);
//    	}
//    }
    static boolean pa(String s,int i,int j)
    {
    	while(i<j)if(s.charAt(i++)!=s.charAt(j--))return false;
    	return true;
    }
    static int ncr(int n,int c,long m)
    {
        long a=1l;
        for(int x=n-c+1;x<=n;x++)a=((a*x)%m);
        long b=1l;
        for(int x=2;x<=c;x++)b=((b*x)%m);
        return (int)((a*(div((int)b,m-2,m)%m))%m);
    }
    static boolean[] sieve(int n)
    {
        boolean bo[]=new boolean[n+1];
        bo[0]=true;bo[1]=true;
        for(int x=4;x<=n;x+=2)bo[x]=true;
        for(int x=3;x*x<=n;x+=2)if(!bo[x])for(int y=x*x;y<=n;y+=x)bo[y]=true;
        return bo;
    }
    static int[] fac(int n)
    {
        int bo[]=new int[n+1];
        for(int x=1;x<=n;x++)for(int y=x;y<=n;y+=x)bo[y]++;
        return bo;
    }
    static long div(long a,long b,long m) 
    {
        long r=1l;
        a%=m;  
        while(b>0)
        {
            if((b&1)==1)r=(r*a)%m; 
            b>>=1;
            a=(a*a)%m;  
        } 
        return r; 
    }
    static int i()throws IOException
    {
        if(!st.hasMoreTokens()) st=new StringTokenizer(br.readLine());
        return Integer.parseInt(st.nextToken());
    }
    static long l()throws IOException
    {
        if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
        return Long.parseLong(st.nextToken());
    }
    static String s()throws IOException
    {
        if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
        return st.nextToken();
    }
    static double d()throws IOException
    {
        if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
        return Double.parseDouble(st.nextToken());
    }
    static void p(Object p){System.out.print(p);}
    static void p(String p){System.out.print(p);}
    static void p(int p){System.out.print(p);}
    static void p(double p){System.out.print(p);}
    static void p(long p){System.out.print(p);}
    static void p(char p){System.out.print(p);}
    static void p(boolean p){System.out.print(p);}
    static void pl(Object p){System.out.println(p);}
    static void pl(String p){System.out.println(p);}
    static void pl(int p){System.out.println(p);}
    static void pl(char p){System.out.println(p);}
    static void pl(double p){System.out.println(p);}
    static void pl(long p){System.out.println(p);}
    static void pl(boolean p){System.out.println(p);}
    static void pl(){System.out.println();}
    static int[] ari(int n)throws IOException
    {
        int ar[]=new int[n];
        if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
        for(int x=0;x<n;x++)ar[x]=Integer.parseInt(st.nextToken());
        return ar;
    }
    static int[][] ari(int n,int m)throws IOException
    {
        int ar[][]=new int[n][m];
        for(int x=0;x<n;x++)
        {
        	if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
            for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());
        }
        return ar;
    }
    static long[] arl(int n)throws IOException
    {
        long ar[]=new long[n];
        if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
        for(int x=0;x<n;x++) ar[x]=Long.parseLong(st.nextToken());
        return ar;
    }
    static long[][] arl(int n,int m)throws IOException
    {
        long ar[][]=new long[n][m];
        for(int x=0;x<n;x++)
        {
        	if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
            for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());
        }
        return ar;
    }
    static String[] ars(int n)throws IOException
    {
        String ar[]=new String[n];
        if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
        for(int x=0;x<n;x++) ar[x]=st.nextToken();
        return ar;
    }
    static double[] ard(int n)throws IOException
    {
        double ar[]=new double[n];
        if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
        for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());
        return ar;
    }
    static double[][] ard(int n,int m)throws IOException
    {
        double ar[][]=new double[n][m];
        for(int x=0;x<n;x++)
        {
        	if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
            for(int y=0;y<m;y++)ar[x][y]=Double.parseDouble(st.nextToken());
        }
        return ar;
    }
    static char[] arc(int n)throws IOException
    {
        char ar[]=new char[n];
        if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
        for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);
        return ar;
    }
    static char[][] arc(int n,int m)throws IOException
    {
        char ar[][]=new char[n][m];
        for(int x=0;x<n;x++)
        {
            String s=br.readLine();
            for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);
        }
        return ar;
    }
    static void p(int ar[])
    {
        StringBuilder sb=new StringBuilder(11*ar.length);
        for(int a:ar)
        {
        	sb.append(a);
        	sb.append(' ');
        }
        System.out.println(sb);
    }
    static void p(int ar[][])
    {
        StringBuilder sb;
        for(int a[]:ar)
        {
            sb=new StringBuilder(2*a.length);
            for(int aa:a)
            {
            	sb.append(aa);
            	sb.append(' ');
            }
            System.out.println(sb);
        }
    }
    static void p(long ar[])
    {
        StringBuilder sb=new StringBuilder(20*ar.length);
        for(long a:ar)
        {
        	sb.append(a);
        	sb.append(' ');
        }
        System.out.println(sb);
    }
    static void p(long ar[][])
    {
        StringBuilder sb;
        for(long a[]:ar)
        {
            sb=new StringBuilder(2*a.length);
            for(long aa:a)
            {
            	sb.append(aa);
            	sb.append(' ');
            }
            System.out.println(sb);
        }
    }
    static void p(String ar[])
    {
        StringBuilder sb=new StringBuilder(2*ar.length);
        for(String a:ar)
        {
        	sb.append(a);
        	sb.append(' ');
        }
        System.out.println(sb);
    }
    static void p(double ar[])
    {
        StringBuilder sb=new StringBuilder(20*ar.length);
        for(double a:ar)
        {
        	sb.append(a);
        	sb.append(' ');
        }
        System.out.println(sb);
    }
    static void p(double ar[][])
    {
        StringBuilder sb;
        for(double a[]:ar)
        {
            sb=new StringBuilder(4*a.length);
            for(double aa:a)
            {
            	sb.append(aa);
            	sb.append(' ');
            }
            System.out.println(sb);
        }
    }
    static void p(char ar[])
    {
        StringBuilder sb=new StringBuilder(2*ar.length);
        for(char aa:ar)
        {
        	sb.append(aa);
        	sb.append(' ');
        }
        System.out.println(sb);
    }
    static void p(char ar[][])
    {
        StringBuilder sb;
        for(char a[]:ar)
        {
            sb=new StringBuilder(2*a.length);
            for(char aa:a)
            {
            	sb.append(aa);
            	sb.append(' ');
            }
            System.out.println(sb);
        }
    }
}