human
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].
Solve the task in PYTHON3.
 Give me a Python code to find the resulting array after performing all the queries?
assistant
#Here is Python code to find the resulting array after performing all the queries.

import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def main(t):

    q = int(input())
    ori = []
    op = []
    groups = []



        
    for _ in range(q):
        temp = list(map(int,input().split()))
        if len(temp)==2:
            ori.append(temp[1])
        else:
            op.append([len(ori),temp[1],temp[2]])


    op = op[::-1]
    
    dic = {}

    ans = []

    def change(e,r):
        if r not in dic:  dic[e] = r
        else:  dic[e] = dic[r]


#    print(ori,dic)       



    index = 0
    for i in range(len(ori)-1,-1,-1):
        while index<len(op) and op[index][0]>i:  
            change(op[index][1],op[index][2])
            index += 1

#        print(dic)
        if ori[i] not in dic:  ans.append(ori[i])
        else:    ans.append( dic[ori[i]]    )
#        print(ans)
    ans = ans[::-1]

        
    









    print(*ans)













T = 1 #int(input())
t = 1
while t<=T:
    main(t)
    t += 1