#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long res = 1ll;
  while (b > 0) {
    if (b % 2ll) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2ll;
  }
  return res;
}
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) {
    res = (res * i) % 1000000007;
  }
  return res % 1000000007;
}
bool comp(pair<string, long long> a, pair<string, long long> b) {
  if (a.second != b.second) return a.second > b.second;
  return a.first < b.first;
}
void solve() {
  long long x = 4;
  long long arr[x];
  for (long long i = 0; i < x; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + x);
  cout << arr[x - 1] - arr[0] << " " << arr[x - 1] - arr[2] << " "
       << arr[x - 1] - arr[1];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  while (tc--) {
    solve();
  }
}

### Prompt

Please formulate 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 power(long long a, long long b) {
  long long res = 1ll;
  while (b > 0) {
    if (b % 2ll) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2ll;
  }
  return res;
}
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) {
    res = (res * i) % 1000000007;
  }
  return res % 1000000007;
}
bool comp(pair<string, long long> a, pair<string, long long> b) {
  if (a.second != b.second) return a.second > b.second;
  return a.first < b.first;
}
void solve() {
  long long x = 4;
  long long arr[x];
  for (long long i = 0; i < x; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + x);
  cout << arr[x - 1] - arr[0] << " " << arr[x - 1] - arr[2] << " "
       << arr[x - 1] - arr[1];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  while (tc--) {
    solve();
  }
}

```