Problem: During the lunch break all n Berland State University students lined up in the food court. However, it turned out that the food court, too, has a lunch break and it temporarily stopped working.

Standing in a queue that isn't being served is so boring! So, each of the students wrote down the number of the student ID of the student that stands in line directly in front of him, and the student that stands in line directly behind him. If no one stands before or after a student (that is, he is the first one or the last one), then he writes down number 0 instead (in Berland State University student IDs are numerated from 1).

After that, all the students went about their business. When they returned, they found out that restoring the queue is not such an easy task.

Help the students to restore the state of the queue by the numbers of the student ID's of their neighbors in the queue.

Input

The first line contains integer n (2 ≤ n ≤ 2·105) — the number of students in the queue. 

Then n lines follow, i-th line contains the pair of integers ai, bi (0 ≤ ai, bi ≤ 106), where ai is the ID number of a person in front of a student and bi is the ID number of a person behind a student. The lines are given in the arbitrary order. Value 0 is given instead of a neighbor's ID number if the neighbor doesn't exist.

The ID numbers of all students are distinct. It is guaranteed that the records correspond too the queue where all the students stand in some order.

Output

Print a sequence of n integers x1, x2, ..., xn — the sequence of ID numbers of all the students in the order they go in the queue from the first student to the last one.

Examples

Input

4
92 31
0 7
31 0
7 141


Output

92 7 31 141 

Note

The picture illustrates the queue for the first sample.

<image>

Solution (in PYTHON):

Solution:
#coding=utf-8
# import numpy as np
'''
思路：
明显站在奇数位置上的人记录了前后两个偶数位置的人  站在偶数位置上的人同样
所以对于偶数位置  只要从0往后找就可以了
对于奇数位置  只要先找到第一个人  然后往后找就可以了
'''
y = {}
def get_even(n,a,b,queue):
    index = 1
    num = a[0]
    while index <n:
        queue[index] = num
        y[num] = 0
        try:
            num = a[num]
        except:
            num = 0
        index += 2
    return queue

def get_odd(n,a,b,queue):
    search_num = 0  # 将要开始查找的id
    for num in y:
        if y[num]  !=0:
            search_num = num
            break
    while num in b:
        num = b[num]
    index = 0
    while index < n :
        queue[index] = num
        try:
            num = a[num]
        except:
            num = 0
        index += 2
    return queue

def data_collect():
    r_i = raw_input()
    r_i = r_i.strip()
    n = int(r_i)
    global y # 记录所有出现的id
    a = {}  # 记录所有前面那个人对应后面人
    b = {}  # 记录所有后面人对应前面人
    for i in range(n):
        r_i = raw_input()
        r_i = r_i.strip().split(' ')
        line_list = [int(t) for t in r_i]
        a[line_list[0]] = line_list[1]
        b[line_list[1]] = line_list[0]
        y[line_list[0]] = 1
    y[0] = 0
    return (n,a,b,y)

def output_print(queue):
    out = list(queue)
    print int(out[0]),
    for i in range(1,len(out)):
        print ' ' + str(int(out[i])),


def main():
    (n,a,b,y) = data_collect()
    queue = []
    for i in range(n):
        queue.append(0)
    get_even(n,a,b,queue)
    get_odd(n,a,b,queue)
    output_print(queue)
    return

main()