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 javafx.util.*; 
import java.math.*;
//import java.lang.*;
public class Main 
{ 
    
   // static int n;
//  static HashSet<Integer> adj[];

   // static int dist[];
   // static int remove[];
//    static boolean isLeaf[];
  //   static Long dp[][];
   // static int arr[];
  //  static boolean ok;
    // static long arr[];
    // static long sum[];
  //  static boolean ok;
    static long mod=1000000007;
 //   static HashSet<Integer> adjacency[];
    static final long oo=(long)1e18;
    public static void main(String[] args) throws IOException { 
    //    Scanner sc=new Scanner(System.in);
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        br = new BufferedReader(new InputStreamReader(System.in));
        int test=nextInt();
      //int test=1;
        
        outer: while(test--!=0){
            int n=nextInt();
            int k=nextInt();
            int arr[][]=new int[n][n];
            int idx=0;
        ou:    while(k>0){
                int j=idx;
                for(int i=0;i<n;i++){
                    arr[i][idx]=1;
                    k--;
                    if(k==0)
                        break ou;
                    idx=(idx+1)%n;
                }
                idx++;
            }
            int ans=0;int max_row=0;int min_row=Integer.MAX_VALUE;int max_col=0;int min_col=Integer.MAX_VALUE;
            for(int i=0;i<n;i++){
                int sum=0;int sum_col=0;
                for(int j=0;j<n;j++){
                    sum+=arr[i][j];
                    sum_col+=arr[j][i];

                }
                if(sum>max_row){
                    max_row=sum;
                }
                if(sum<min_row){
                    min_row=sum;
                }
                if(sum_col>max_col){
                    max_col=sum_col;
                }
                if(sum_col<min_col){
                    min_col=sum_col;
                }
                
            }
            ans=(max_row-min_row)*(max_row-min_row)+(max_col-min_col)*(max_col-min_col);
            pw.println(ans);
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    pw.print(arr[i][j]);
                }
                pw.println();
            }

        }
           
        pw.close();
    }

    static class Pair implements Comparable<Pair>{
        double g,ai,bi,sa,sb;
        Pair(double g,double ai,double bi){
          //  this.index=index;
            this.g=g;
            this.ai=ai;
            this.bi=bi;
            sa=g/ai;
            sb=g/bi;
            //this.z=z;
        }   
        public int compareTo(Pair p){
            double a=Math.abs(sa-sb);
            double b=Math.abs(p.sa-p.sb);
            if(a>b){
                return 1;
            }
            else if(a==b){
                return 0;

            }
            else{
                return -1;
            }
        }
    }
   static void update(long tree[],int idx,long val){
        //println(idx);
        while(idx<1000006){
            
            tree[idx]+=val;
            idx+=(idx)&(-idx);
        }
    }   
    static long sum(long tree[],int idx){
        long sum=0;
        while(idx>0){
           
            sum+=tree[idx];
            idx-=(idx)&(-idx);
        }
        return sum;
    }
    public static BufferedReader br;
    public static StringTokenizer st;
    public static String next() {
        while (st == null || !st.hasMoreTokens()) {
       try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }
 
    public static Integer nextInt() {
        return Integer.parseInt(next());
    }
    public static Long nextLong() {
        return Long.parseLong(next());
    }
 
    public static Double nextDouble() {
        return Double.parseDouble(next());
    }
    // static class Pair{
    //     int x;int y;
    //     Pair(int x,int y,int z){
    //         this.x=x;
    //         this.y=y;
    //        // this.z=z;
    //      //   this.z=z;
    //       //  this.i=i;
    //     }
    // }
    // static class sorting implements Comparator<Pair>{
    //     public int compare(Pair a,Pair b){
    //         //return (a.y)-(b.y);
    //         if(a.y==b.y){
    //             return -1*(a.z-b.z);
    //         }
    //         return (a.y-b.y);
    //     }
   // }

    static long ncr(long n,long r){
        
        if(r==0)
            return 1l;
        long val=ncr(n-1,r-1);
        val=(n*val);
        val=(val/r);
        return val;
    }

    public static int[] na(int n)throws IOException{
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = nextInt();
        return a;
    }
    static class query implements Comparable<query>{
        int l,r,idx,block;
        static int len;
        query(int l,int r,int i){
            this.l=l;
            this.r=r;
            this.idx=i;
            this.block=l/len;
        }   
        public int compareTo(query a){
            return block==a.block?r-a.r:block-a.block;
        }
    }
    
    // static class sorting implements Comparator<Pair>{  
    //     public int compare(Pair a1,Pair a2){  
    //         if(o1.a==o2.a)
    //             return (o1.b>o2.b)?1:-1;  
    //         else if(o1.a>o2.a)  
    //             return 1;  
    //         else  
    //             return -1;  
    //     }
    // }  
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static int gcd(int a, int b) { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    // To compute x^y under modulo m 
    static long power(long x, long y, long m){ 
        if (y == 0) 
            return 1;      
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    }
    static long fast_pow(long base,long n,long M){
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
       static long modInverse(long n,long M){
        return fast_pow(n,M-2,M);
    }
    // (1,1)   

    //     (3,2)   (3,5)

}