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 PYTHON3):

Solution:
#------------------------------what is this I don't know....just makes my mess faster--------------------------------------
import os
import sys
from io import BytesIO, IOBase
 
BUFSIZE = 8192
 
class FastIO(IOBase):
	newlines = 0
 
	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None
 
	def read(self):
		while True:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			if not b:
				break
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines = 0
		return self.buffer.read()
 
	def readline(self):
		while self.newlines == 0:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			self.newlines = b.count(b"\n") + (not b)
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines -= 1
		return self.buffer.readline()
 
	def flush(self):
		if self.writable:
			os.write(self._fd, self.buffer.getvalue())
			self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		self.write = lambda s: self.buffer.write(s.encode("ascii"))
		self.read = lambda: self.buffer.read().decode("ascii")
		self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
#----------------------------------Real game starts here--------------------------------------

#_______________________________________________________________#
def fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x-1)
def lower_bound(li, num): #return 0 if all are greater or equal to
	answer = -1 
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer #index where x is not less than num
def upper_bound(li, num): #return n-1 if all are small or equal
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2

		if li[middle] <= num:
			answer = middle
			start = middle + 1
		
		else:
			end = middle - 1
	return answer #index where x is not greater than num

def abs(x):
	return x if x >=0 else -x
def binary_search(li, val, lb, ub):
	ans = 0
	while(lb <= ub):
		mid = (lb+ub)//2
		#print(mid, li[mid])
		if li[mid] > val:
			ub = mid-1
		elif val > li[mid]:
			lb = mid + 1
		else:
			ans = 1
			break
	return ans

def sieve_of_eratosthenes(n):
	ans = []
	arr = [1]*(n+1)
	arr[0],arr[1], i = 0, 0, 2
	while(i*i <= n):
		if arr[i] == 1:
			j = i+i
			while(j <= n):
				arr[j] = 0
				j += i
		i += 1
	for k in range(n):
		if arr[k] == 1:
			ans.append(k)
	return ans
def nc2(x):
	if x == 1:
		return 0
	else:
		return x*(x-1)//2
def kadane(x):
	sum_so_far = 0
	current_sum = 0
	for i in x:
		current_sum += i
		if current_sum < 0:
			current_sum = 0
		else:
			sum_so_far = max(sum_so_far,current_sum)
	return sum_so_far

#_______________________________________________________________#
'''
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'''
from math import *
#for _ in range(1):
for _ in range(int(input())):
	#n = int(input())
	n,k = map(int,input().split())
	mat = [[0 for i in range(n)] for i in range(n)]
	c = -1
	for i in range(k):
		if i%n == 0:
			c += 1
		mat[i%n][c%n] = 1
		c += 1
	print(2 if k%n else 0)
	for i in range(n):
		print(''.join(str(x) for x in mat[i]))