#include <bits/stdc++.h>
using namespace std;
long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long binary_expo(long long base, long long expo) {
  long long res = 1;
  while (expo) {
    if (expo % 2) {
      res = mul(res, base);
    }
    expo /= 2;
    base = mul(base, base);
  }
  return res;
}
void solve() {
  long long A[5];
  for (long long i = 1; i <= 4; i++) {
    cin >> A[i];
  }
  sort(A + 1, A + 1 + 4);
  long long maxx = A[4];
  do {
    long long bb = (A[1] - A[2] + A[3]) / 2;
    long long cc = A[3] - bb;
    long long aa = A[2] - cc;
    if (aa >= 0 and bb >= 0 and cc >= 0 and aa + bb + cc == maxx) {
      cout << aa << " " << bb << " " << cc << endl;
      return;
    }
  } while (next_permutation(A + 1, A + 1 + 3));
}
signed main() {
  long long t;
  ios_base::sync_with_stdio(false);
  t = 1;
  while (t--) {
    solve();
  }
  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>
using namespace std;
long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long binary_expo(long long base, long long expo) {
  long long res = 1;
  while (expo) {
    if (expo % 2) {
      res = mul(res, base);
    }
    expo /= 2;
    base = mul(base, base);
  }
  return res;
}
void solve() {
  long long A[5];
  for (long long i = 1; i <= 4; i++) {
    cin >> A[i];
  }
  sort(A + 1, A + 1 + 4);
  long long maxx = A[4];
  do {
    long long bb = (A[1] - A[2] + A[3]) / 2;
    long long cc = A[3] - bb;
    long long aa = A[2] - cc;
    if (aa >= 0 and bb >= 0 and cc >= 0 and aa + bb + cc == maxx) {
      cout << aa << " " << bb << " " << cc << endl;
      return;
    }
  } while (next_permutation(A + 1, A + 1 + 3));
}
signed main() {
  long long t;
  ios_base::sync_with_stdio(false);
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```