Let's denote the Manhattan distance between two points $p_1$ (with coordinates $(x_1, y_1)$) and $p_2$ (with coordinates $(x_2, y_2)$) as $d(p_1, p_2) = |x_1 - x_2| + |y_1 - y_2|$. For example, the distance between two points with coordinates $(1, 3)$ and $(4, 2)$ is $|1 - 4| + |3 - 2| = 4$.

You are given two points, $A$ and $B$. The point $A$ has coordinates $(0, 0)$, the point $B$ has coordinates $(x, y)$.

Your goal is to find a point $C$ such that:

both coordinates of $C$ are non-negative integers;

$d(A, C) = \dfrac{d(A, B)}{2}$ (without any rounding);

$d(B, C) = \dfrac{d(A, B)}{2}$ (without any rounding).

Find any point $C$ that meets these constraints, or report that no such point exists.


-----Input-----

The first line contains one integer $t$ ($1 \le t \le 3000$) — the number of test cases.

Each test case consists of one line containing two integers $x$ and $y$ ($0 \le x, y \le 50$) — the coordinates of the point $B$.


-----Output-----

For each test case, print the answer on a separate line as follows:

if it is impossible to find a point $C$ meeting the constraints, print "-1 -1" (without quotes);

otherwise, print two non-negative integers not exceeding $10^6$ — the coordinates of point $C$ meeting the constraints. If there are multiple answers, print any of them. It can be shown that if any such point exists, it's possible to find a point with coordinates not exceeding $10^6$ that meets the constraints.


-----Examples-----

Input
10
49 3
2 50
13 0
0 41
42 0
0 36
13 37
42 16
42 13
0 0
Output
23 3
1 25
-1 -1
-1 -1
21 0
0 18
13 12
25 4
-1 -1
0 0


-----Note-----

Explanations for some of the test cases from the example:

In the first test case, the point $B$ has coordinates $(49, 3)$. If the point $C$ has coordinates $(23, 3)$, then the distance from $A$ to $B$ is $|49 - 0| + |3 - 0| = 52$, the distance from $A$ to $C$ is $|23 - 0| + |3 - 0| = 26$, and the distance from $B$ to $C$ is $|23 - 49| + |3 - 3| = 26$.

In the second test case, the point $B$ has coordinates $(2, 50)$. If the point $C$ has coordinates $(1, 25)$, then the distance from $A$ to $B$ is $|2 - 0| + |50 - 0| = 52$, the distance from $A$ to $C$ is $|1 - 0| + |25 - 0| = 26$, and the distance from $B$ to $C$ is $|1 - 2| + |25 - 50| = 26$.

In the third and the fourth test cases, it can be shown that no point with integer coordinates meets the constraints.

In the fifth test case, the point $B$ has coordinates $(42, 0)$. If the point $C$ has coordinates $(21, 0)$, then the distance from $A$ to $B$ is $|42 - 0| + |0 - 0| = 42$, the distance from $A$ to $C$ is $|21 - 0| + |0 - 0| = 21$, and the distance from $B$ to $C$ is $|21 - 42| + |0 - 0| = 21$.

Difficulty: EASY
Tags: Complete search

Constructive algorithms
Skill types: Complete search

Starter code:

Solutions:
def solution(b_x, b_y):
	if (b_x + b_y) % 2:
		return '-1 -1'
	distance = (b_x + b_y) // 2
	if distance < b_x:
		return f'{distance} 0'
	return f'{b_x} {distance - b_x}'
print(*(solution(*map(int, input().split())) for _ in range(int(input()))), sep='\n')


import sys
input = sys.stdin.readline

def solve():
	(x, y) = map(int, input().split())
	if (x + y) % 2 == 1:
		return [-1, -1]
	if (x + y) // 2 > x:
		return [x, (x + y) // 2 - x]
	return [(x + y) // 2, 0]
for _ in range(int(input())):
	print(*solve())


t = int(input())
for j in range(t):
	(x, y) = map(int, input().split())
	if (x + y) % 2 == 1:
		print(-1, -1)
	elif x < y:
		print(x, y - (x + y) // 2)
	else:
		print((x + y) // 2, 0)


for i in range(int(input())):
	(a, b) = [int(x) for x in input().split()]
	s = a + b
	if s % 2 != 0:
		print(-1, -1)
	else:
		s //= 2
		val = min(a, s)
		print(val, max(s - val, 0))


def distance(x, y):
	a = x + y
	if a % 2 == 0:
		a //= 2
		print(min(a, x), max(a - x, 0))
	else:
		print('-1 -1')
in_t_case = int(input())
for i in range(in_t_case):
	(x, y) = map(int, input().split())
	distance(x, y)