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Bitwise exclusive OR (or bitwise addition modulo two) is a binary operation which is equivalent to applying logical exclusive OR to every pair of bits located on the same positions in binary notation of operands. In other words, a binary digit of the result is equal to 1 if and only if bits on the respective positions in the operands are different.

For example, if X = 109₁₀ = 1101101₂, Y = 41₁₀ = 101001₂, then:

X xor Y = 68₁₀ = 1000100₂.

Write a program, which takes two non-negative integers A and B as an input and finds two non-negative integers X and Y, which satisfy the following conditions:

A = X + Y
B = X xor Y, where xor is bitwise exclusive or.
X is the smallest number among all numbers for which the first two conditions are true.

Input

The first line contains integer number A and the second line contains integer number B (0 ≤ A, B ≤ 2⁶⁴ - 1).

Output

The only output line should contain two integer non-negative numbers X and Y. Print the only number -1 if there is no answer.

Examples

Input

142
76

Output

33 109

题目大意：

给定的两个数a,b,满足x+y=a,x^y=b,输出x最小的一组x,y

思路：

因为x^y是不进位的加法

所以 A的二进制中的进位得到的1 可以由 A-B 得到

然后若整除2 那么就有解，解就是 (A-B)/2

若不能则无解

#include <iostream>
#include <cmath>
#include <cstdio>

using namespace std;

typedef long long ll;

//a = x + y 
//b = x ^ y


int main() {
	ll a,b;
	
	cin >> a >> b;
	ll x = (a - b) / 2;
	ll y = x + b;
    
    	
    if(x + y == a) {
        printf("%lld %lld\n",x,y);
    } else {
        printf("-1\n");
    }
    
	return 0;
}