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 Today {
	static void solve() throws IOException {
		int n = nextInt(), k = nextInt(), kk = k, ans[][] = new int[n][n], dev = 0, row, col, right = 0;
		int x = 0, y = 0;
		for(int i = 0; i < k; i++) {
			if(i % n == 0) {
				x = 0;
				y = right;
				right++;
			}
			ans[x][y] = 1;
			x = (x + 1) % n;
			y = (y + 1) % n;
		}
		out.println((kk % n == 0) ? 0 : 2);
		for(int i = 0; i < n; ++i) {
			for(int j = 0; j < n; ++j) {
				out.print(ans[i][j]);
			}
			out.println();
		}
	}

	public static void main(String args[]) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(new BufferedOutputStream(System.out));
		// sieve();
		int tt = nextInt();
		// int tt = 1;
		for(int test = 1; test <= tt; ++test) {
			// out.print("Case #" + test + ": ");
			solve();
		}
		out.close();
	}

	static class Pair implements Comparable<Pair> {
		long first, second;

		Pair(long a, long b) {
			first = a;
			second = b;
		}

		public int compareTo(Pair p) {
			return (int) (this.first - p.first);
		}

		public boolean equals(Object p) {
			Pair p1 = (Pair) p;
			return (first == p1.first && second == p1.second);
		}

		public String toString() {
			return this.first + " " + this.second;
		}

		public int hashCode() {
			return (int) ((1l * (inf + 1) * this.first + this.second) % mod);
		}
	}

	static boolean arraySortedOrNot(int a[], int n) {
		if (n == 1 || n == 0)
			return true;
		if (a[n - 1] < a[n - 2])
			return false;
		return arraySortedOrNot(a, n - 1);
	}

	static void swap(int a[], int i, int j) {
		int temp = a[i];
		a[i] = a[j];
		a[j] = temp;
	}

	static long binaryExpo(long a, long b) {
		if(a == 1)
			return 1;
		long res = 1;
		a = a % mod;
		while(b != 0) {
			if((b & 1) != 0) {
				res = (res % mod * a % mod) % mod;
			}
			a = (a % mod * a % mod) % mod;
			b >>= 1;
		}
		return res;
	}

	static boolean isPerfectSquare(long x) {
		double sr = Math.sqrt(x);
		return ((sr - Math.floor(sr)) == 0);
	}

	static long isPowerOfTwo(long n) {
		if (n == 0)
			return -1;
		long count = 0;

		while (n != 1) {
			++count;
			if (n % 2 != 0)
				return -1;
			n = n / 2;
		}
		return count;
	}

	static final long mod = (long) (1e9 + 7);
	static final int inf = (int) (1e9 + 1);

	public static int lowerBound(int[] array, int length, int 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;
	}

	public static int upperBound(int[] array, int length, int 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;
	}

	static int gcd(int a, int b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static BufferedReader br;
	static StringTokenizer st;
	static PrintWriter out;

	static String next() {
		while (st == null || !st.hasMoreElements()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	static int nextInt() {
		return Integer.parseInt(next());
	}

	static long nextLong() {
		return Long.parseLong(next());
	}

	static double nextDouble() {
		return Double.parseDouble(next());
	}

	static String nextLine() {
		String str = "";
		try {
			str = br.readLine();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return str;
	}

	static int[] nextIntArray(int n) {
		int a[] = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}
		return a;
	}

	static long[] nextLongArray(int n) {
		long a[] = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextLong();
		}
		return a;
	}

	static int[] memset(int n, int val) {
		int ar[] = new int[n];
		Arrays.fill(ar, val);
		return ar;
	}

	static void debug(Object... a) {
		System.out.print("> ");
		for (int i = 0; i < a.length; i++)
			System.out.print(a[i] + " ");
		System.out.println();
	}

	static void debug(int a[]) {
		debugnsp(Arrays.stream(a).boxed().toArray());
	}

	static void debug(long a[]) {
		debugnsp(Arrays.stream(a).boxed().toArray());
	}

	static void debugnsp(Object a[]) {
		System.out.print("> ");
		for (int i = 0; i < a.length; i++)
			System.out.print(a[i] + " ");
		System.out.println();
	}

	// static final int MAXN = 100001;
	// static int spf[] = new int[MAXN];

	// static void sieve() {
	// 	spf[1] = 1;
	// 	for (int i = 2; i < MAXN; i++)
	// 		spf[i] = i;
	// 	for (int i = 4; i < MAXN; i += 2)
	// 		spf[i] = 2;

	// 	for (int i = 3; i * i < MAXN; i++) {
	// 		if (spf[i] == i) {
	// 			for (int j = i * i; j < MAXN; j += i)
	// 				if (spf[j] == j)
	// 					spf[j] = i;
	// 		}
	// 	}
	// }
}

/*
 * Jai Sita Ram Ji
 *
 * @author: Nishchal Siddharth Pandey
 */