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].
1
1 50
 1
1 114514
 1
1 2002
 1
1 500

50 
 114514 
 2002 
 500 

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define int long long
#define watch(x) cerr << (#x) << " is " << (x) << endl;
#define endl '\n'
#define all(x) (x).begin(), (x).end()
#define sz(a) (int)(a).size()
#define vi vector<int>
#define pii pair<int, int>
#define fi first
#define se second
const int maxn = (int)5e5+500;

int n;
int a[maxn];
int nxt[maxn];
map<int, vi> mp;
signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int q;
    cin >> q;

    int cur=0;
    while(q--) {
        int t;
        cin >> t;

        if(t==1) {
            int x;
            cin >> x;

            a[++cur] = x;
            mp[x].emplace_back(cur);
        } else {
            int x,y;
            cin >> x >> y;
            if(mp[x].empty()) continue;
            int rep = mp[x].back();
            for(int idx: mp[x]) {
                nxt[idx] = rep;
            }
            a[rep] = y;
            mp[y].emplace_back(rep);

            mp[x].clear();
        }
    }

    for(int i = cur; i >= 1; i--) {
        if(nxt[i]) a[i] = a[nxt[i]];
    }

    for(int i = 1; i <= cur; i++) {
        cout << a[i] << " ";
    }
    cout << endl;
}

/*

dsu thing

*/