A

Doubly linked list is one of the fundamental data structures. A doubly linked list is a sequence of elements, each containing information about the previous and the next elements of the list. In this problem all lists have linear structure. I.e. each element except the first has exactly one previous element, each element except the last has exactly one next element. The list is not closed in a cycle.In this problem you are given n memory cells forming one or more doubly linked lists. Each cell contains information about element from some list. Memory cells are numbered from 1 to n.For each cell i you are given two values:    li — cell containing previous element for the element in the cell i;    ri — cell containing next element for the element in the cell i.If cell i contains information about the element which has no previous element then li = 0. Similarly, if cell i contains information about the element which has no next element then ri = 0.Three lists are shown on the picture.

For example, for the picture above the values of l and r are the following: l1 = 4, r1 = 7; l2 = 5, r2 = 0; l3 = 0, r3 = 0; l4 = 6, r4 = 1; l5 = 0, r5 = 2; l6 = 0, r6 = 4; l7 = 1, r7 = 0.

Your task is to unite all given lists in a single list, joining them to each other in any order. In particular, if the input data already contains a single list, then there is no need to perform any actions. Print the resulting list in the form of values li, ri.Any other action, other than joining the beginning of one list to the end of another, can not be performed.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of memory cells where the doubly linked lists are located.Each of the following n lines contains two integers li, ri (0 ≤ li, ri ≤ n) — the cells of the previous and the next element of list for cell i. Value li = 0 if element in cell i has no previous element in its list. Value ri = 0 if element in cell i has no next element in its list.It is guaranteed that the input contains the correct description of a single or more doubly linked lists. All lists have linear structure: each element of list except the first has exactly one previous element; each element of list except the last has exactly one next element. Each memory cell contains information about one element from some list, each element of each list written in one of n given cells.

Output

Print n lines, the i-th line must contain two integers li and ri — the cells of the previous and the next element of list for cell i after all lists from the input are united in a single list. If there are many solutions print any of them.

Example
Input

74 75 00 06 10 20 41 0Output4 75 60 56 13 22 41 0

题意：这个题就是说输入n个数前面的数和后面的数，然后根据前后关系把他们连成一条链，如果有多个链则合并成一条。在合并过程中使用并查集很方便（使用并查集根据他们前面 的数连成链）。合并的条件为这个数的l = 0，则为链首，可以合并在r = 0的后面（即链尾）

#include<stdio.h>#include<stdlib.h>#include<string.h>struct Node{    int front;    int next;} a[104];int f[110];void init(int n){    for(int i = 1; i <= n; i++)    {        f[i] = i;    }    return;}int getf(int v){    if(v == f[v])///当f[v] = v的时候说明他是链首，是头领。    return v;    else///若不是则递归寻找    {        f[v] = getf(f[v]);        return f[v];    }}int merge_f(int x, int y)///相同头领的进行归并{    int t1 = getf(x);    int t2 = getf(y);    if(t1 != t2)    {        f[t2] = t1;        return 1;    }    else    {        return 0;    }}int main(){    int n, i, j;    while(~scanf("%d", &n))    {        init(n);        for(i = 1; i <= n; i++)        {            scanf("%d %d", &a[i].front, &a[i].next);            if(a[i].front != 0)                f[i] = a[i].front;///i的前一个数为他的头领，记在f[i]里        }        for(i = 1; i <= n; i++)        {            if(a[i].next != 0)                continue;            for(j = 1; j <= n; j++)            {                if(i != j && a[j].front == 0)                {                    if(merge_f(i, j))                    {                        a[i].next = j;                        a[j].front = i;                        break;                    }                }            }        }        for(i = 1; i <= n; i++)        {            printf("%d %d\n", a[i].front, a[i].next);        }    }    return 0;}