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.(话说题意理解错是怎么过前三个样例的...) 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.

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.

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.

The picture illustrates the queue for the first sample.

附代码（当然也是人家的。0。0）

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 using namespace std; 5 const int maxn = 1000010; 6 int stu[maxn][2], l[maxn], r[maxn]; 7 int line[maxn]; 8 int main() 9 {10     //freopen("in.txt", "r", stdin);11     int n;12     scanf("%d", &n);13     for(int i = 0; i < n; i++)14     {15         scanf("%d%d", &l[i], &r[i]);16         stu[r[i]][0] = l[i];17         stu[l[i]][1] = r[i];18     }19     int j, i;20     /*找出偶数位的人，并将确定位置的人的值设为-1*/21     for(i = stu[0][1], j = 2; i; j+=2)22     {23         line[j] = i;24         int tmp = stu[i][1];25         stu[i][0] = -1;    stu[i][1] = -1;26         i = tmp;27     }28     /*根据前前位为0找第一个奇数; 开始此处不理解，其实就是在定义的时候已经把第一个奇数的前前位设为0了,而第一个偶数的前前位经上面步骤变为-1,故不会引起冲突*/29     for(int k = 0; k < n; k++)30     {31         if(!stu[l[k]][0])   {i = l[k];  break;}32     }33     /*找奇数*/34     for(j = 1; i; j+=2)35     {36         line[j] = i;37         i = stu[i][1];38     }39     for(int l = 1; l <= n; l++) {if(l != 1) printf(" ");    printf("%d", line[l]);}40     printf("\n");41     return 0;42 }

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难道只能当一题党吗。