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 Main {
 
	public static void main(String[] args) throws IOException,InterruptedException{
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			int k=sc.nextInt();
			int[][] arr=new int[n][n];
			int a=k/n;
			int b=k%n;		
			boolean[] freq=new boolean[n]; 		
			for (int i = 0; i < n; i++) {
				for (int j = i; j < a+i; j++) {
					int ss=0;
					if(j>=n) ss=j-n;
					else ss=j;
					arr[i][ss]=1;
				}
			}
			for (int i = 0; i < n; i++) {
				if(b<=0) break;
				for (int j = 0; j < n; j++) {
					if(arr[i][j]==0&&!freq[j]) {
						arr[i][j]=1;
						freq[j]=true;
						b--;
						break;
					}
				}
			}
			long maxc=0,minc=1000000000,maxr=0,minr=1000000000;
			for (int i = 0; i < arr.length; i++) {
				int c=0;
				for (int j = 0; j < arr.length; j++) {
					c+=arr[i][j];
				}
				minr=Math.min(minr,c);
				maxr=Math.max(maxr,c);
			}
			for (int i = 0; i < arr.length; i++) {
				int c=0;
				for (int j = 0; j < arr.length; j++) {
					c+=arr[j][i];
				}
				minc=Math.min(minc,c);
				maxc=Math.max(maxc,c);
			}
//			pw.println(maxr+" "+minr);
			long s=(maxr-minr)*(maxr-minr)+(maxc-minc)*(maxc-minc);
			pw.println(s);
			for (int i = 0; i < arr.length; i++) {
				for (int j = 0; j < arr.length; j++) {
					pw.print(arr[i][j]);
				}
				pw.println();
			}
		}
		pw.close();
    }
	
	static PrintWriter pw=new PrintWriter(System.out);
	static long pow(int a,int b) {
		long r=1l;
		for (int i = 0; i < b; i++) {
			r*=a;
		}
		return r;
	}
	static boolean isprime(long n) {
		for (int i = 2; i <= Math.sqrt(n); i++) {
			if(n%i==0) return false;
		}
		return true;
	}
	static int[]lp;
	static void sieveLinear(int N){
		ArrayList<Integer> primes = new ArrayList<Integer>();
		lp = new int[N + 1];								//lp[i] = least prime divisor of i
		for(int i = 2; i <= N; ++i){
			if(lp[i] == 0){
				primes.add(i);
				lp[i] = i;
			}
			int curLP = lp[i];
			for(int p: primes)//all primes smaller than or equal my lowest prime divisor
				if(p > curLP || p * 1l * i > N)
					break;
				else
					lp[p * i] = p;
		}
	}
	static long gcd(int x,int y) {
		while (x!=y) {
			if(Math.max(x,y)/Math.min(x,y)==(double)(Math.max(x,y))/Math.min(x,y)) return Math.min(x,y);
			if(lp.length!=0) {
				if(lp[x]==x) {
					if(y/x==y/(double)x) return x;
					else return 1;
				}else if (lp[y]==y) {
					if(x/y==x/(double)y) return y;
					else return 1;
				}	
			}
			if(x>y) x-=y;
			else y-=x;
		}
		return x;
	}
	static class pair implements Comparable<pair> {
		int x;
		int y;
 
		public pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
 
		public String toString() {
			return x + " " + y;
		}
		public boolean equals(Object o) {
            if (o instanceof pair) {
                pair p = (pair)o;
                return p.x == x && p.y == y;
            }
            return false;
        }
        public int hashCode() {
            return new Double(x).hashCode() * 31 + new Double(y).hashCode();
        }
        public int compareTo(pair other) {
        	if(this.x==other.x) {
        		return Long.compare(this.y, other.y);
        	}
			return Long.compare(this.x, other.x);
		}
	}
	static class tuble implements Comparable<tuble> {
		int x;
		int y;
		int z;
 
		public tuble(int x, int y, int z) {
			this.x = x;
			this.y = y;
			this.z = z;
		}
 
		public String toString() {
			return x + " " + y + " " + z;
		}
 
		public int compareTo(tuble other) {
			if (this.x == other.x) {
				if(this.y==other.y) return this.z - other.z;
				else return this.y - other.y;
			} else {
				return this.x - other.x;
			}
		}
	}
 	static class Scanner {
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}
 
		public boolean hasNext() {
			// TODO Auto-generated method stub
			return false;
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
	}
 
}