#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using P = pair<int, int>;
#define PI acos(-1)
#define Mod (int)1000000007
#define INFTY (int)INT_MAX
#define Rep(i, n) for (int i = 0; i < (int)(n); i++)
#define BitRep(i, n) for (int i = 0; i < (int)(1 << n); i++)
#define All(vec) vec.begin(), vec.end()
#define Part(vec, b, e) vec.begin() + b, vec.begin() + e
#define Sum(vec) accumulate(vec.begin(), vec.end(), 0)
#define Sort(vec) sort(vec.begin(), vec.end())
#define Reverse(vec) reverse(vec.begin(), vec.end())
#define Count(vec, x) count(vec.begin(), vec.end(), x)
#define Next_permutation(vec) next_permutation(vec.begin(), vec.end())
#define Find(S, s) S.find(s) != string::npos

int main() {
	int n;
	cin >> n;
	vector<int> A(n);
	Rep(i, n) cin >> A[i];
	A.erase(unique(All(A)), A.end());
	Rep(i, A.size()) {
		if (i) cout << " ";
		cout << A[i];
	}
	cout << endl;
	return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

For a sequence of integers $A = \\{a_0, a_1, ..., a_{n-1}\\}$ which is sorted by ascending order, eliminate all equivalent elements.

Constraints

* $1 \leq n \leq 100,000$
* $-1000,000,000 \leq a_i \leq 1,000,000,000$
* $a_0 \leq a_1 \leq ... \leq a_{n-1}$

Input

A sequence is given in the following format.


$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$


Output

Print the sequence after eliminating equivalent elements in a line. Separate adjacency elements by a space character.

Example

Input

4
1 2 2 4


Output

1 2 4

### Response

```cpp
#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using P = pair<int, int>;
#define PI acos(-1)
#define Mod (int)1000000007
#define INFTY (int)INT_MAX
#define Rep(i, n) for (int i = 0; i < (int)(n); i++)
#define BitRep(i, n) for (int i = 0; i < (int)(1 << n); i++)
#define All(vec) vec.begin(), vec.end()
#define Part(vec, b, e) vec.begin() + b, vec.begin() + e
#define Sum(vec) accumulate(vec.begin(), vec.end(), 0)
#define Sort(vec) sort(vec.begin(), vec.end())
#define Reverse(vec) reverse(vec.begin(), vec.end())
#define Count(vec, x) count(vec.begin(), vec.end(), x)
#define Next_permutation(vec) next_permutation(vec.begin(), vec.end())
#define Find(S, s) S.find(s) != string::npos

int main() {
	int n;
	cin >> n;
	vector<int> A(n);
	Rep(i, n) cin >> A[i];
	A.erase(unique(All(A)), A.end());
	Rep(i, A.size()) {
		if (i) cout << " ";
		cout << A[i];
	}
	cout << endl;
	return 0;
}

```