#include <bits/stdc++.h>
const long long int MOD = 1e9 + 7;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
using namespace std;
void vok() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n == 5 or n == 7) return true;
  if (n % 2 == 0 or n % 5 == 0 or n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i += 6) {
    if (n % i == 0 or n % (i + 2) == 0) return false;
  }
  return true;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return ((a * b) / gcd(a, b));
}
vector<long long int> pp;
void gen() {
  vector<bool> prime(33000, true);
  prime[0] = prime[1] = false;
  for (int p = 0; p * p <= 33000; p++) {
    if (prime[p]) {
      for (int i = p * p; i <= 33000; i += p) prime[i] = false;
    }
  }
  for (int i = 0; i < 33000; i++)
    if (prime[i]) pp.push_back(i);
}
long long int find(vector<int>& A, int n) {}
void solve() {
  long long int n, m, x, y, a, b, c, d, k;
  long long int mx = -1;
  long long int mi = INT_MAX;
  a = b = x = y = k = c = 0;
  long long int ans = 0, res = 0;
  vector<int> dp;
  cin >> a >> b >> c >> d;
  dp.push_back(a);
  dp.push_back(b);
  dp.push_back(c);
  dp.push_back(d);
  sort(dp.begin(), dp.end());
  cout << dp[3] - dp[2] << " " << dp[3] - dp[1] << " " << dp[3] - dp[0] << endl;
}
int main() {
  vok();
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
const long long int MOD = 1e9 + 7;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
using namespace std;
void vok() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n == 5 or n == 7) return true;
  if (n % 2 == 0 or n % 5 == 0 or n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i += 6) {
    if (n % i == 0 or n % (i + 2) == 0) return false;
  }
  return true;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return ((a * b) / gcd(a, b));
}
vector<long long int> pp;
void gen() {
  vector<bool> prime(33000, true);
  prime[0] = prime[1] = false;
  for (int p = 0; p * p <= 33000; p++) {
    if (prime[p]) {
      for (int i = p * p; i <= 33000; i += p) prime[i] = false;
    }
  }
  for (int i = 0; i < 33000; i++)
    if (prime[i]) pp.push_back(i);
}
long long int find(vector<int>& A, int n) {}
void solve() {
  long long int n, m, x, y, a, b, c, d, k;
  long long int mx = -1;
  long long int mi = INT_MAX;
  a = b = x = y = k = c = 0;
  long long int ans = 0, res = 0;
  vector<int> dp;
  cin >> a >> b >> c >> d;
  dp.push_back(a);
  dp.push_back(b);
  dp.push_back(c);
  dp.push_back(d);
  sort(dp.begin(), dp.end());
  cout << dp[3] - dp[2] << " " << dp[3] - dp[1] << " " << dp[3] - dp[0] << endl;
}
int main() {
  vok();
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```