Farmer John owns a large number of fences that must be repaired annually. He traverses the fences by riding a horse along each and every one of them (and nowhere else) and fixing the broken parts.

Farmer John is as lazy as the next farmer and hates to ride the same fence twice. Your program must read in a description of a network of fences and tell Farmer John a path to traverse each fence length exactly once, if possible. Farmer J can, if he wishes, start and finish at any fence intersection.

Every fence connects two fence intersections, which are numbered inclusively from 1 through 500 (though some farms have far fewer than 500 intersections). Any number of fences (>=1) can meet at a fence intersection. It is always possible to ride from any fence to any other fence (i.e., all fences are "connected").

Your program must output the path of intersections that, if interpreted as a base 500 number, would have the smallest magnitude.

There will always be at least one solution for each set of input data supplied to your program for testing.

PROGRAM NAME: fence

INPUT FORMAT

SAMPLE INPUT (file fence.in)

9
1 2
2 3
3 4
4 2
4 5
2 5
5 6
5 7
4 6

OUTPUT FORMAT

The output consists of F+1 lines, each containing a single integer. Print the number of the starting intersection on the first line, the next intersection's number on the next line, and so on, until the final intersection on the last line. There might be many possible answers to any given input set, but only one is ordered correctly.

SAMPLE OUTPUT (file fence.out)

1
2
3
4
2
5
4
6
5
7

代碼：

/*
LANG: C
TASK: fence
*/
#include<stdio.h>
#define maxn 501
#define Max(a, b) (a > b ? a : b)
#define Min(a, b) (a < b ? a : b)
int num[maxn], map[maxn][maxn], min, max, path[maxn], index = 0;
void circuit(int x)
{
    int i;
    if (num[x])
    {
        do
        {
            for (i = min; i <= max; i++)
            {
                if (map[x][i])//找最小的鄰接點
                {
                    break;
                }
            }
            map[x][i]--, map[i][x]--, num[x]--, num[i]--;//刪除該邊 
            circuit(i);//搜索x的鄰接點 
        }while (num[x]);
        path[index++] = x;//搜索點x的所有鄰接點后才保存路勁x
    }
    else
    {
        path[index++] = x;
    }
}
void deal()
{
    int i;
    for (i = min; i <= max; i++)//如果有度為奇數的點，從奇數點搜索 
    {
        if (num[i] % 2 == 1)
        {
            circuit(i);
            return;
        }
    }
    circuit(min);//否則，任意一點都可以，但題意要求歐拉回路盡可能從小到大的順序輸出 
}
void print()
{
    int i;
    for (i = index - 1; i >= 0; i--) 
    {
        printf("%d\n", path[i]);
    }
}
void read()
{
    int i, n, u, v;
    scanf("%d", &n);
    for (i = 0, min = 500, max = 1 ; i < n; i++)
    {
        scanf("%d%d", &u, &v);
        map[u][v]++, map[v][u]++;
        num[u]++, num[v]++;
        if (max < Max(u, v))
        {
            max = Max(u, v);
        }
        if (min > Min(u, v))
        {
            min = Min(u, v);
        }
    }
}        
int main()
{
    freopen("fence.in", "r", stdin), freopen("fence.out", "w", stdout);
    read();
    deal();
    print();
    fclose(stdin), fclose(stdout);
    //system("pause");
    return 0;
}