#include <bits/stdc++.h>
int main() {
  int a, b, c, d, e, j, k, l;
  scanf("%d %d %d %d", &a, &b, &c, &d);
  if (a > b && a > c && a > d) {
    e = a;
    j = a - b;
    k = a - c;
    l = a - d;
  } else if (b > c && b > a && b > d) {
    e = b;
    j = b - a;
    k = b - c;
    l = b - d;
  } else if (c > a && c > b && c > d) {
    e = c;
    j = c - b;
    k = c - a;
    l = c - d;
  } else {
    e = d;
    j = d - b;
    k = d - c;
    l = d - a;
  }
  printf("%d %d %d  ", j, k, l);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  int a, b, c, d, e, j, k, l;
  scanf("%d %d %d %d", &a, &b, &c, &d);
  if (a > b && a > c && a > d) {
    e = a;
    j = a - b;
    k = a - c;
    l = a - d;
  } else if (b > c && b > a && b > d) {
    e = b;
    j = b - a;
    k = b - c;
    l = b - d;
  } else if (c > a && c > b && c > d) {
    e = c;
    j = c - b;
    k = c - a;
    l = c - d;
  } else {
    e = d;
    j = d - b;
    k = d - c;
    l = d - a;
  }
  printf("%d %d %d  ", j, k, l);
  return 0;
}

```