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.*;
public class MyClass {
    
    
    public static void main(String args[]) {
      FastReader sc = new FastReader();     //For Fast IO
      //func f = new func();                  //Call func for swap , permute, upper bound and for sort.
      
     int t = sc.nextInt();
     while(t-->0){
         int n = sc.nextInt();
         int k = sc.nextInt();
         int arr[][] = new int[n][n];
         int f = (k%n==0)?0:2;
         for(int i=0;i<n&&k>0;i++){
             int x=0;
             int y=i;
             for(int j=0;j<n&&k>0;j++){
                 arr[x][y]=1;
                 x=(x+1)%n;
                 y=(y+1)%n;
                 k--;
             }
         }
         System.out.println(f);
         for(int i=0;i<n;i++){
             for(int j=0;j<n;j++){
                 System.out.print(arr[i][j]+"");
             }
             System.out.println();
         }
     }
      
      
      
      
    }
}
class Pair{
    int x;
    int y;
    Pair(int i,int j){
        x=i;
        y=j;
    }
    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair key = (Pair) o;
        return x == key.x && y == key.y;
    }
    @Override
    public int hashCode() {
        int result = x;
        result = 31*result + y;
        return result;
    }
}
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 func{
    static ArrayList<String> al = new ArrayList<>();
        public static void sort(long[] arr) {
            int n = arr.length, mid, h, s, l, i, j, k;
            long[] res = new long[n];
            for (s = 1; s < n; s <<= 1) {
                for (l = 0; l < n - 1; l += (s << 1)) {
                    h = Math.min(l + (s << 1) - 1, n - 1);
                    mid = Math.min(l + s - 1, n - 1);
                    i = l;
                    j = mid + 1;
                    k = l;
                    while (i <= mid && j <= h) res[k++] = (arr[i] <= arr[j] ? arr[i++] : arr[j++]);
                    while (i <= mid) res[k++] = arr[i++];
                    while (j <= h) res[k++] = arr[j++];
                    for (k = l; k <= h; k++) arr[k] = res[k];
                }
            }
        }
        public static void permute(char a[] , int i , int n){
            if(i==n-1){
                String s = new String(a);
                al.add(s);                      // al stores all permutations of string. 
                return;
            }
            for(int j=i;j<n;j++){
                swap(a,i,j);
                permute(a,i+1,n);
                swap(a,i,j);
            }
        }
        public static void swap(char a[],int i, int j){
            char temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
        public static void swap(int a[],int i, int j){
            int temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
		//-It returns index of first element which is strictly greater than searched value.
		public static int upperBound(long[] array, int length, long value) {
            int low = 0;
            int high = length;
            while (low < high) {
                final int mid = (low + high) / 2;
                if (value >= array[mid]) {
                    low = mid + 1;
                } else {
                    high = mid;
                }
        }
        return low;
    }
	/*If searched element doesn't exist function returns index of first element which is bigger than searched value.<br>
 * -If searched element is bigger than any array element function returns first index after last element.<br>
 * -If searched element is lower than any array element function returns index of first element.<br>
 * -If there are many values equals searched value function returns first occurrence.<br>*/
	public static int lowerBound(long[] array, int length, long value) {
        int low = 0;
        int high = length;
        while (low < high) {
            final int mid = (low + high) / 2;
            if (value <= array[mid]) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low;
    }
        
 
}