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 PYTHON):

Solution:
# template by 3xC and starkizard.
# contributors: 

#####################################################################################
from __future__ import division, print_function 

import sys
import os
from collections import Counter, deque, defaultdict
import itertools
import math
import io


"""Uncomment modules according to your need"""


# from bisect import bisect_left, bisect_right, insort
# from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
# from copy import deepcopy
# from decimal import *
# import re
# import operator

#####################################################################################
# this enables you to write python3 code with PyPy2 (Python 2)
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

#####################################################################################

"""value of mod"""
MOD = 998244353
mod = 10**9 + 7

"""Uncomment next 4 lines if doing huge recursion"""
# import threading
# threading.stack_size(1<<27)
# sys.setrecursionlimit(10000)

def prepare_factorial(mod=mod):
    """ returns two lists, factorial and inverse factorial modulo argument by default 10**9 +7 """
    # Comment code out when you don't need inverse factorial or vice versa
    fact = [1]
    for i in range(1, 200005):
        fact.append((fact[-1] * i) % mod)

    ifact = [0] * 200005
    ifact[200004] = pow(fact[200004], mod - 2, mod)
    for i in range(200004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod

    return fact, ifact

def modinv(n, p):
    """ returns N inverse modulo p """
    return pow(n, p - 2, p)

def ncr(n, r,  fact, ifact,MOD=mod):
    """ takes 4 arguments: n , r and factorial and inverse factorial lists"""
    t = (fact[n] * (ifact[r]*ifact[n-r]) % MOD)% MOD
    return t


def get_n(Sum):    
    """this function returns the maximum n for which Summation(n) <= Sum"""
    ans = (-1 + sqrt(1 + 8*Sum))//2
    return ans

def sieve(n):
    """ returns a list of prime numbers till n """
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r



def divs(n, start=1):
    """ returns a list of all divisors till n """
    divisors = []
    #rdivisors=[]
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors



def divn(n, primes):
    """ returns the number of divisors, two arguments n and the sieve till n """
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
    return divs_number


def lrfind(d, x, default=-1):
    """ Takes 2 arguments an iterable and an element. returns a tuple (firstoccurence,lastoccurence) -1 if not found """
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return default, default
    else:
        return left, right

def gcd(x, y): # math.gcd is slower
    """ returns greatest common divisor of x and y """
    while y:
        x, y = y, x % y
    return x

def check_sorted(a):
    ''' returns True/False '''
    for i in range(len(a)-1):
        if a[i]>a[i+1]:
            return False
    return True

def print_arr2d(a):
    for i in a:
        prr(i)
    
def ceil(n, k=1): return n // k + (n % k != 0) #returns math.ceil but protecting against floating inconsistencies
def input(): return sys.stdin.readline().strip()
def ii(): return int(input()) #inputs integer
def mi(): return map(int, input().split()) # inputting space seperated variables for example x,y,z
def li(): return list(map(int, input().split())) #inputting a space seperated list of integers
def lw(): return input().split() #inputting a space seperated list of strings
def lcm(a, b): return abs(a * b) // gcd(a, b) #returns LCM of two arguments
def prr(a, sep=' ', end='\n'): print(sep.join(map(str, a)), end=end) #For printing an iterable with seperator sep as optional second argument (default : " "), ending character (default: "\n") as optional third
def dd(): return defaultdict(int) #returns a dictionary with values defaulted to 0
def ddl(): return defaultdict(list) #returns a dictionary with values defaulted to []
def write(s): return sys.stdout.write(s)












###################################################################
# code goes here
def main():
    for t in range(ii()):
        n,k=mi()
        matrix=[]
        init=k//n
        for i in range(n):
            row=[0 for j in range(n)]
            for x in range(init):
                row[(i+x)%n]=1
            matrix.append(row)
        rem=k-(init*n)
        for i in range(rem):
            index=(i+init)%n
            matrix[i][index]=1
        if rem==0:
            print(0)
        else:
            print(2)
        for i in matrix:
            for j in i:
                print(j,end="")
            print()




            


            
                
            

            



                






        
        


    
    
                

    

""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

""" main function"""

if __name__ == '__main__':
    main()
    #threading.Thread(target=main).start()