Almost Ternary Matrix
You are given two even integers $$$n$$$ and $$$m$$$. Your task is to find any binary matrix $$$a$$$ with $$$n$$$ rows and $$$m$$$ columns where every cell $$$(i,j)$$$ has exactly two neighbours with a different value than $$$a_{i,j}$$$.Two cells in the matrix are considered neighbours if and only if they share a side. More formally, the neighbours of cell $$$(x,y)$$$ are: $$$(x-1,y)$$$, $$$(x,y+1)$$$, $$$(x+1,y)$$$ and $$$(x,y-1)$$$.It can be proven that under the given constraints, an answer always exists.
White means $$$0$$$, black means $$$1$$$. The binary matrix from the first test caseThe binary matrix from the second test caseThe binary matrix from the third test case 
Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The following lines contain the descriptions of the test cases.The only line of each test case contains two even integers $$$n$$$ and $$$m$$$ ($$$2 \le n,m \le 50$$$) — the height and width of the binary matrix, respectively.
For each test case, print $$$n$$$ lines, each of which contains $$$m$$$ numbers, equal to $$$0$$$ or $$$1$$$ — any binary matrix which satisfies the constraints described in the statement.It can be proven that under the given constraints, an answer always exists.
t = int(input())
for _ in range(t):
    (n, m) = input().split(' ')
    (n, m) = (int(n), int(m))
    (s1, s2) = ('', '')
    for i in range(m):
        if i % 4 == 0 or i % 4 == 3:
            s1 = s1 + '1 '
            s2 = s2 + '0 '
        else:
            s1 = s1 + '0 '
            s2 = s2 + '1 '
    for i in range(n):
        if i % 4 == 0 or i % 4 == 3:
            print(s1)
        else:
            print(s2)
t = int(input())
while t > 0:
    t -= 1
    (n, m) = map(int, input().split())
    s1 = '1'
    while True:
        s1 += '0'
        if len(s1) == m:
            break
        s1 += '0'
        if len(s1) == m:
            break
        s1 += '1'
        if len(s1) == m:
            break
        s1 += '1'
        if len(s1) == m:
            break
    s2 = ''
    for i in range(m):
        if s1[i] == '1':
            s2 += '0'
        else:
            s2 += '1'
    s1 = ' '.join(s1)
    s2 = ' '.join(s2)
    print(s1)
    n -= 1
    while True:
        if n == 0:
            break
        n -= 1
        print(s2)
        if n == 0:
            break
        n -= 1
        print(s2)
        if n == 0:
            break
        n -= 1
        print(s1)
        if n == 0:
            break
        n -= 1
        print(s1)
for i in range(int(input())):
    (n, m) = map(int, input().split())
    org = '0 1 '
    rev = '1 0 '
    first = ''
    second = ''
    for i in range(m // 2):
        if i % 2:
            first += rev
            second += org
        else:
            first += org
            second += rev
    for i in range(n // 2):
        if i % 2:
            print(first)
            print(second)
        else:
            print(second)
            print(first)