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 ≤ - 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
题意：
给出两整数A,B，要求找到X,Y满足以下三个条件：
1.A=X+Y
2.B=X xor Y2其中xor表示异或运算
3.X是所有满足前两个条件中最小的
思路：
异或是不进位的加法，X和Y做异或运算，那么，他们在同一个 位置的1会变成0，而其他一个1一个0的位置，则会变成1，因此做异或运算会比加法运算少2倍的相同位置的1的值，因此的值就是X和Y相同的部分，而X要求最小并且又必须要有相同部分，因此X可以取得最小就可以是，两者的相同部分，其他的都由Y来提供。
再看不成立的条件，首先当A小于B时是不成立的，因为异或是不进位的加法如果两个数没有1在同一个位置上，那么异或的值会和加法值相同，否则会有1变为0，异或的结果会变小。其次A-B是奇数的时候也是不成立的，由上面分析我们可以得出，A-B必定是偶数。
因为这道题的数据到了-1超过longlong的取值范围，而在unsigned long long范围内，因此可以用unsigned long long。
longlong取值范围为-2^63到2^63-1，大概为9x10^18这么多。
#include<stdio.h>
#include<string.h>
#include<queue>
#include<set>
#include<iostream>
#include<map>
#include<stack>
#include<cmath>
#include<algorithm>
#define ll long long
#define mod 1000000007
#define eps 1e-8
using namespace std;
int main()
{
unsigned ll a,b;
cin>>a>>b;
unsigned ll tt,y,x;
tt=a-b;
if(b>a||tt&1)
{
printf("-1\n");
return 0;
}
x=tt>>1;
y=a-x;
cout<<x<<" "<<y<<endl;
return 0;
}