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.util.logging.Logger;

import static java.lang.System.*;

public class A1371 {
    static BufferedReader br;
    static  long mod  = 998244353;
    static HashSet<Integer> p  = new HashSet<>();
    public static void main(String[] args) throws Exception {
        br = new BufferedReader(new InputStreamReader(System.in));
        // System.out.println((rem*i));
        int testcCase = 1;
        testcCase = cinI();
        while (testcCase-- > 0) {
           // System.out.println(testcCase +" ffff ");
             int[] nk  = readArray(2,0,0);
             int n =(nk[0]);int k = nk[1];
             int[][] g = new int[n][n];
          int sum = 0 ;


                 int p = k / n; int last  = k%n;





            int m  =0;
         // System.out.println(p +" "+last);
                 for(int r=0;r<n;r++){
                     int x =last<=0?0:1;
                     int times = p +x;
                    // System.out.println(r+"  "+times);
                     int  i = m;
                     for(int j=i; times>0;j++,times--){
                         g[r][j%n]=1;
                        // System.out.println(r +" "+j);
                        // System.out.println(g[r][j]+" r "+r+" j "+j);
                         sum+=1;
                         if(sum>k){
                             break;
                         }
                         m=(j+1)%n;
                     }
                     last-=1;
                     if(sum>k){break;}
                 }
                 int max= 0;
                 int min = Integer.MAX_VALUE;
                 for(int u=0;u<n;u++){
                     int sum1 = 0;
                     for(int   l = 0; l<n ;l++){
                         sum1+=g[u][l];
                     }
                     max=max(sum1,max);
                     min=min(min,sum1);
                 }
         // System.out.println(max + " max min "+min);
                 int ans = (max-min)*(max-min);
                 max=0;min=Integer.MAX_VALUE;
                 for(int u = 0 ;u<n;u++){                     int sum1 = 0;

                     for(int  l =0;l<n ;l++){
                         sum1+=g[l][u];

                     }
                     max=max(sum1,max);
                     min=min(min,sum1);
                 }
                 ans = ans+ ((max-min)*(max-min));
          //  System.out.println(max + " max col min "+min);
            System.out.println(ans);
                 for(int   u = 0; u<n;u++){
                     StringBuilder  ob = new StringBuilder();
                     for(int l = 0; l< n ;l++){
                         ob.append(g[u][l]);
                     }
                     System.out.println(ob.toString());

                 }








        }

    }







    public static long min(long a,long b) {
        return Math.min(a,b);
    }
    public static int min(int a,int b) {
        return Math.min(a,b);
    }





























    public  static  int[] readArray(int n,int x ,int z)throws Exception{
        int[] arr= new int[n];
        String[] ar= cinA();
        for(int i =x ;i<n+x;i++){
            arr[i]=getI(ar[i-x]);
        }
        return arr;
    }
    public  static  long[] readArray(int n,int x )throws Exception{
        long[] arr= new long[n];
        String[] ar= cinA();
        for(int i =x ;i<n+x;i++){
            arr[i]=getL(ar[i-x]);
        }
        return arr;
    }

    public static void arrinit(String[] a,long[] b)throws Exception{
        for(int i=0;i<a.length;i++){
            b[i]=Long.parseLong(a[i]);
        }
    }
    public  static HashSet<Integer>[]  Graph(int n,int edge,int directed)throws Exception{
        HashSet<Integer>[] tree=  new HashSet[n];

        for(int j=0;j<edge;j++){

            String[] uv = cinA();
            int u = getI(uv[0]);
            int v = getI(uv[1]);
            if(directed==0){

                tree[v].add(u);
            }
            tree[u].add(v);
        }
        return tree;
    }
    public static void arrinit(String[] a,int[] b)throws Exception{
        for(int i=0;i<a.length;i++){
            b[i]=Integer.parseInt(a[i]);
        }
    }


    static double findRoots(int a, int b, int c) {
        // If a is 0, then equation is not
        //quadratic, but linear


        int d = b * b - 4 * a * c;
        double sqrt_val = Math.sqrt(Math.abs(d));


        // System.out.println("Roots are real and different \n");

        return Math.max((double) (-b + sqrt_val) / (2 * a),
                (double) (-b - sqrt_val) / (2 * a));


    }

    public static String cin() throws Exception {
        return br.readLine();
    }

    public static String[] cinA() throws Exception {
        return br.readLine().split(" ");
    }
    public static String[] cinA(int x) throws Exception{
        return br.readLine().split("");
    }

    public static String ToString(Long x) {
        return Long.toBinaryString(x);
    }

    public static void cout(String s) {
        System.out.println(s);
    }

    public static Integer cinI() throws Exception {
        return Integer.parseInt(br.readLine());
    }

    public static int getI(String s) throws Exception {
        return Integer.parseInt(s);
    }

    public static long getL(String s) throws Exception {
        return Long.parseLong(s);
    }

    public static long max(long a, long b) {
        return Math.max(a, b);
    }
    public static int max(int a, int b) {
        return Math.max(a, b);
    }

    public static void coutI(int x) {
        System.out.println(String.valueOf(x));
    }

    public static void coutI(long x) {
        System.out.println(String.valueOf(x));
    }

    public static Long cinL() throws Exception {
        return Long.parseLong(br.readLine());
    }

    public static void arrInit(String[] arr, int[] arr1) throws Exception {
        for (int i = 0; i < arr.length; i++) {
            arr1[i] = getI(arr[i]);
        }

    }
}