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 powTwo(long long n) {
  if (n > 1000000000) {
    long long ans = 0;
    while (n > 1) {
      ans++;
      n /= 2;
    }
    return (ans);
  } else {
    double ans = log10(n) * 1.0 / log10(2) * 1.0;
    return (ans);
  }
}
string getBinary(long long n) {
  string bin = bitset<32>(n).to_string();
  long long cp;
  string s;
  for (long long i = 0; i < 32; i++) {
    if (bin[i] == '1') {
      cp = i;
      break;
    }
  }
  for (long long i = cp; i < 32; i++) {
    s += bin[i];
  }
  return (s);
}
float power(float x, int y) {
  float temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else {
    if (y > 0)
      return x * temp * temp;
    else
      return (temp * temp) / x;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  vector<long long> l(4);
  cin >> l[0] >> l[1] >> l[2] >> l[3];
  sort(l.begin(), l.end());
  cout << l[3] - l[0] << " " << l[3] - l[1] << " " << l[3] - l[2] << endl;
  return (0);
}