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].
impor
Write 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.
t sys
input = sys.stdin.readline

class Node():
    def __init__(self,val):
        self.val = val
        self.next = None

n = int(input())
cnt = 0
d = {}
g = {}
for i in range(n):
    l = list(map(int,input().split()))
    if len(l) == 2:
        x = l[1]
        now = Node(cnt)
        if x in d and d[x] != None:
            d[x].next = now
            d[x] = d[x].next
            
        else:
            d[x] = now
            g[x] = now
            
        cnt += 1
        
    else:
        x,y = l[1],l[2]
        if x not in d or d[x] == None:
            continue
        
        if x == y:
            continue
        
        if y in d and d[y] != None:
            d[y].next = g[x]
            d[y] = d[x]
            
        else:
            d[y] = d[x]
            g[y] = g[x]
            
        d[x] = None
        g[x] = None
            
ans = [0 for i in range(cnt)]
for x in g:
    while g[x] != None:
        ans[g[x].val] = x
        g[x] = g[x].next
        
print(*ans)