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.io.*;
 
public class Mamo {
    static long mod=998244353;
    static Reader in=new Reader();
    static List<Integer >G[];
    static boolean B=true;
    static StringBuilder Sl=new StringBuilder();
    static int DP[][];
    public static void main(String [] args) {
    	//Dir by MohammedElkady
    int t=in.nextInt();
    	while(t-->0) {
    		int n=in.nextInt(),k=in.nextInt();
    		int div=k/n;
    		int a[][]=new int [n][n];
    		if(k%n==0) {out.append(0+"\n");}
    		else {out.append(2+"\n");}
    		for(int i=0;i<n;i++) {
    			int l=0;
    			if(k%n!=0) {l++;k--;}
    			for(int u=i%n,o=div+l;u<n+i&&o>0;u++,o--) {
    				a[i][u%n]=1;}}
    		for(int i=0;i<n;i++) {
    			for(int u=0;u<n;u++) {
    				out.append(a[i][u]+"");
    			}
    			out.append("\n");}
    	}
    out.close();}

    
    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static long gcd(long g,long x){if(x<1)return g;else return gcd(x,g%x);}
    static class Reader 
    { 
        private InputStream mIs;private byte[] buf = new byte[1024];private int curChar,numChars;public Reader() { this(System.in); }public Reader(InputStream is) { mIs = is;} 
        public int read() {if (numChars == -1) throw new InputMismatchException();if (curChar >= numChars) {curChar = 0;try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}if (numChars <= 0) return -1; }return buf[curChar++];} 
        public String nextLine(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isEndOfLine(c));return res.toString() ;} 
        public String s(){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 long l(){int c = read();while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }long 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 int nextInt(){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 double d() throws IOException {return Double.parseDouble(s()) ;}
        public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } 
        public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } 
        public int[] arr(int n){int[] ret = new int[n];for (int i = 0; i < n; i++) {ret[i] = nextInt();}return ret;}
    }
}
class node implements Comparable<node>{
int f, a,b;
node(int tt,int ll,int vv){
f=tt;a=ll;b=vv;
}
    @Override
    public int compareTo(node o) {
    return f-o.f;    
    }
}
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;
        }
}
 
class Sorting{
	 
	 public static int[] bucketSort(int[] array, int bucketCount) {
	 
	 if (bucketCount <= 0) throw new IllegalArgumentException("Invalid bucket count");
	 
	 if (array.length <= 1) return array; //trivially sorted
	 
	 
	 
	 int high = array[0];
	 
	 int low = array[0];
	 
	 for (int i = 1; i < array.length; i++) { //find the range of input elements
	 
	 if (array[i] > high) high = array[i];
	 
	 if (array[i] < low) low = array[i];
	 
	 }
	 
	 double interval = ((double)(high - low + 1))/bucketCount; //range of one bucket
	 
	 
	 ArrayList<Integer> buckets[] = new ArrayList[bucketCount];
	 
	 for (int i = 0; i < bucketCount; i++) { //initialize buckets
	 
	 buckets[i] = new ArrayList();
	 
	 }
	 
	 
	 
	 for (int i = 0; i < array.length; i++) { //partition the input array
	 
	 buckets[(int)((array[i] - low)/interval)].add(array[i]);
	 
	 }
	 
	 
	 int pointer = 0;
	 
	 for (int i = 0; i < buckets.length; i++) {
	 
	 Collections.sort(buckets[i]); //mergeSort
	 
	 for (int j = 0; j < buckets[i].size(); j++) { //merge the buckets
	 
	 array[pointer] = buckets[i].get(j);
	 
	 pointer++;
	 
	 }
	 
	}
	 
	return array;
	 
	}
	}