題幹：

Superbowl Sunday is nearly here. In order to pass the time waiting for the half-time commercials and wardrobe malfunctions, the local hackers have organized a betting pool on the game. Members place their bets on the sum of the two final scores, or on the absolute difference between the two scores. 
Given the winning numbers for each type of bet, can you deduce the final scores? 

Input

The first line of input contains n, the number of test cases. n lines follow, each representing a test case. Each test case gives s and d, non-negative integers representing the sum and (absolute) difference between the two final scores.

Output

For each test case, output a line giving the two final scores, largest first. If there are no such scores, output a line containing "impossible". Recall that football scores are always non-negative integers.

Sample Input

2
40 20
20 40

Sample Output

30 10
impossible

題目大意：

   就是說有倆數x和y，現在告訴你    這兩個數的和a、這兩個數的差的絕對值b。  問你能否構造出這兩個數，如果沒有符合條件的解，輸出-1，如果有，就從大到小輸出x和y。

解題報告：

   水題啊，找個規律，發現如果加和為奇數的話，，肯定是不滿足的啊（證明我也沒細想，反正寫出來之後小範圍測試了一下沒啥問題就交了）

AC程式碼：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
ll a,b;
int main()
{
	int t;
	cin>>t;
	while(t--) {
		scanf("%lld%lld",&a,&b);
		if(a<b) {
			puts("impossible");continue;
		}
		ll ans1 = (a+b)>>1;
		if(ans1*2 != (a+b)) {
			puts("impossible");continue;
		}
		ll ans2 = a-ans1;
		if(ans1 < ans2) swap(ans1,ans2);
		printf("%lld %lld\n",ans1,ans2);
	}
	return 0 ;
 }