Problem: Polycarp has guessed three positive integers a, b and c. 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 x_1, x_2, x_3, x_4 (2 ≤ x_i ≤ 10^9) — numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.

Output

Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int RTN();
int main() {
  RTN();
  return 0;
}
int RTN() {
  int n1, n2, n3, n4, scase = 1, max, min1, min2, min3, x = 1, a, b, c;
  cin >> n1;
  cin >> n2;
  cin >> n3;
  cin >> n4;
  switch (scase) {
    case 1:
      if (n1 > n2 && n1 > n3 && n1 > n4) {
        max = n1;
        min1 = n2;
        min2 = n3;
        min3 = n4;
        scase = 2;
        break;
      }
    case 2:
      if (n2 > n1 && n2 > n3 && n2 > n4) {
        max = n2;
        min1 = n1;
        min2 = n3;
        min3 = n4;
        scase = 3;
        break;
      }
    case 3:
      if (n3 > n1 && n3 > n2 && n3 > n4) {
        max = n3;
        min1 = n1;
        min2 = n2;
        min3 = n4;
        scase = 4;
        break;
      }
    case 4:
      max = n4;
      min1 = n1;
      min2 = n2;
      min3 = n3;
      scase = 0;
      break;
  }
  a = ((min2 - min3) + min1) / 2;
  b = ((min1 - min2) + min3) / 2;
  c = ((min3 - min1) + min2) / 2;
  printf("%d %d %d", a, b, c);
  return 0;
}