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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << "[ " << *it << " = " << a << " ] ";
  err(++it, args...);
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const pair<Tk, Tv>& p) {
  os << "{" << p.first << ',' << p.second << "}";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[ ";
  for (T x : p) os << x << " ";
  os << "]" << '\n';
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}" << '\n';
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const map<Tk, Tv>& p) {
  os << "{ ";
  for (pair<Tk, Tv> x : p) os << x << " ";
  os << "}" << '\n';
  return os;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int a[4];
  for (long long int i = 0; i < 4; i++) {
    cin >> a[i];
  }
  sort(a, a + 4);
  cout << a[3] - a[0] << " " << a[3] - a[1] << " " << a[3] - a[2] << '\n';
  return 0;
}