#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline int qpow(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n >>= 1;
  }
  return ans;
}
inline long long _qpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
inline long long ksm(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
const int maxnum = 5e5 + 5;
struct que {
  int op, x, y;
};
vector<int> ans;
vector<que> v;
int f[maxnum];
bool vis[maxnum];
int op, x, y, q;
void slove() {
  cin >> q;
  while (q--) {
    cin >> op;
    if (op == 1) {
      cin >> x;
      v.push_back({1, x, 0});
    } else {
      cin >> x >> y;
      v.push_back({2, x, y});
    }
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < maxnum; i++) f[i] = i;
  for (auto q : v) {
    op = q.op, x = q.x, y = q.y;
    if (op == 1)
      ans.push_back(f[x]);
    else
      f[x] = f[y];
  }
  reverse(ans.begin(), ans.end());
  for (int x : ans) cout << x << " ";
  cout << "\n";
}
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  ;
  slove();
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

You have an array of integers (initially empty).

You have to perform q queries. Each query is of one of two types: 

  * "1 x" — add the element x to the end of the array; 
  * "2 x y" — replace all occurrences of x in the array with y. 



Find the resulting array after performing all the queries.

Input

The first line contains a single integer q (1 ≤ q ≤ 5 ⋅ 10^5) — the number of queries.

Next q lines contain queries (one per line). Each query is of one of two types: 

  * "1 x" (1 ≤ x ≤ 5 ⋅ 10^5); 
  * "2 x y" (1 ≤ x, y ≤ 5 ⋅ 10^5). 



It's guaranteed that there is at least one query of the first type.

Output

In a single line, print k integers — the resulting array after performing all the queries, where k is the number of queries of the first type.

Examples

Input


7
1 3
1 1
2 1 2
1 2
1 1
1 2
2 1 3


Output


3 2 2 3 2 

Input


4
1 1
1 2
1 1
2 2 2


Output


1 2 1 

Input


8
2 1 4
1 1
1 4
1 2
2 2 4
2 4 3
1 2
2 2 7


Output


1 3 3 7 

Note

In the first example, the array changes as follows:

[] → [3] → [3, 1] → [3, 2] → [3, 2, 2] → [3, 2, 2, 1] → [3, 2, 2, 1, 2] → [3, 2, 2, 3, 2].

In the second example, the array changes as follows:

[] → [1] → [1, 2] → [1, 2, 1] → [1, 2, 1].

In the third example, the array changes as follows:

[] → [] → [1] → [1, 4] → [1, 4, 2] → [1, 4, 4] → [1, 3, 3] → [1, 3, 3, 2] → [1, 3, 3, 7].

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline int qpow(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n >>= 1;
  }
  return ans;
}
inline long long _qpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
inline long long ksm(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
const int maxnum = 5e5 + 5;
struct que {
  int op, x, y;
};
vector<int> ans;
vector<que> v;
int f[maxnum];
bool vis[maxnum];
int op, x, y, q;
void slove() {
  cin >> q;
  while (q--) {
    cin >> op;
    if (op == 1) {
      cin >> x;
      v.push_back({1, x, 0});
    } else {
      cin >> x >> y;
      v.push_back({2, x, y});
    }
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < maxnum; i++) f[i] = i;
  for (auto q : v) {
    op = q.op, x = q.x, y = q.y;
    if (op == 1)
      ans.push_back(f[x]);
    else
      f[x] = f[y];
  }
  reverse(ans.begin(), ans.end());
  for (int x : ans) cout << x << " ";
  cout << "\n";
}
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  ;
  slove();
  return 0;
}

```