Polycarp has guessed three positive integers aa , bb and cc . He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order — their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b , a+c , b+c and a+b+c .
You have to guess three numbers a , b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a , b and c can be equal (it is also possible that a=b=c ).
Input
The only line of the input contains four positive integers x1,x2,x3,x4 (2≤xi≤1092≤xi≤109 ) — numbers written on a board in random order. It is guaranteed that the answer exists for the given number x1,x2,x3,x4x1,x2,x3,x4 .
Output
Print such positive integers aa , bb and cc that four numbers written on a board are values a+ba+b , a+ca+c , b+cb+c and a+b+ca+b+c written in some order. Print aa , bb and cc in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100题意：给你四个数字让你猜他自己想要的三个数字，第一个数字代表a+b，第二个数字代表a+c，第三个数字代表b+c，第四个数字代表a+b+c，我们可以想，首先对这四个数字进行一个排序，然后找到最大的那个通过for循环依次减去前边的数字就是答案，可以想一下。
#include<bits/stdc++.h>
using namespace std;
int main(){
int s[5];
for(int i = 0 ; i < 4 ; i ++){
cin >> s[i];
}
sort(s,s+4);
for(int i = 0 ; i < 3 ; i ++){
printf("%d ",s[3] - s[i]);
}
printf("\n");
}