#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  vector<long long> x(4);
  for (__typeof(0) i = (0); i < (4); i++) cin >> x[i];
  sort(x.rbegin(), x.rend());
  long long all = x[0];
  for (__typeof(1) i = (1); i < (4); i++) {
    for (__typeof(1) j = (1); j < (4); j++) {
      if (i == j) continue;
      for (__typeof(1) k = (1); k < (4); k++) {
        if (k == i || k == j) continue;
        long long c = all - x[i];
        long long b = all - x[j];
        long long a = all - x[k];
        if (a > 0 && b > 0 && c > 0 && a + b + c == all) {
          cout << a << " " << b << " " << c << '\n';
          return 0;
        }
      }
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  vector<long long> x(4);
  for (__typeof(0) i = (0); i < (4); i++) cin >> x[i];
  sort(x.rbegin(), x.rend());
  long long all = x[0];
  for (__typeof(1) i = (1); i < (4); i++) {
    for (__typeof(1) j = (1); j < (4); j++) {
      if (i == j) continue;
      for (__typeof(1) k = (1); k < (4); k++) {
        if (k == i || k == j) continue;
        long long c = all - x[i];
        long long b = all - x[j];
        long long a = all - x[k];
        if (a > 0 && b > 0 && c > 0 && a + b + c == all) {
          cout << a << " " << b << " " << c << '\n';
          return 0;
        }
      }
    }
  }
  return 0;
}

```