#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<ld> vld;
typedef vector<string> vstr;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;  // 小さい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;     // 大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); ++i)
#define TS to_string
#define ALL(v) (v).begin(), (v).end()
#define endl "\n"

ll INF = 1e9;
ll MOD = 1000000007;
ll LINF = 1e18;
ld EPS = 1e-9;
ld PI = M_PI;
vll dx = {1, 0, -1, 0, 1, -1, -1, 1};
vll dy = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b;}     //最小公倍数

void yes(){ cout << "Yes" << endl;}
void no(){ cout << "No" << endl;}

//-----------------------------------------

//-----------------------------------------
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  ll n;
  cin >> n;
  set<ll> s;
  ll x;
  REP(i,n) { cin >> x; s.insert(x);}
  IREP(i,s){
    if(i != s.begin()) cout << " ";
    cout << (*i);
  }
  cout << endl;
  return 0;
}



### Prompt

Please provide a Cpp coded solution to the problem described below:

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;

typedef long long ll;
typedef long double ld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<ld> vld;
typedef vector<string> vstr;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;  // 小さい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;     // 大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); ++i)
#define TS to_string
#define ALL(v) (v).begin(), (v).end()
#define endl "\n"

ll INF = 1e9;
ll MOD = 1000000007;
ll LINF = 1e18;
ld EPS = 1e-9;
ld PI = M_PI;
vll dx = {1, 0, -1, 0, 1, -1, -1, 1};
vll dy = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b;}     //最小公倍数

void yes(){ cout << "Yes" << endl;}
void no(){ cout << "No" << endl;}

//-----------------------------------------

//-----------------------------------------
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  ll n;
  cin >> n;
  set<ll> s;
  ll x;
  REP(i,n) { cin >> x; s.insert(x);}
  IREP(i,s){
    if(i != s.begin()) cout << " ";
    cout << (*i);
  }
  cout << endl;
  return 0;
}



```