As we all kown, MZL hates the endless loop deeply, and he commands you to solve this problem to end the loop.
You are given an undirected graph with n vertexs and m edges. Please direct all the edges so that for every vertex in the graph the inequation |out degree ? in degree|≤1 is satisified.
The graph you are given maybe contains self loops or multiple edges.

The first line of the input is a single integer T, indicating the number of testcases.
For each test case, the first line contains two integers n and m.
And the next m lines, each line contains two integers ui and vi, which describe an edge of the graph.
T≤100, 1≤n≤105, 1≤m≤3?105, ∑n≤2?105, ∑m≤7?105.

For each test case, if there is no solution, print a single line with ?1, otherwise output m lines,.
In ith line contains a integer 1 or 0, 1 for direct the ith edge to ui→vi, 0 for ui←vi.

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

1
1
1
0
1
0
1
0
1

题意: 给一个n个点,m条边的无向图,要求给m条边定方向,使得每个定点的出入度之差的绝对值小于等于1. 输出任意一种结果.(输出m条边的方向)

题解: 
      
     <1>定理: 一个图,肯定存在偶数个奇度点.
     通过<1>,我们可以找出若干条欧拉路径,直接按遍历的路径,定一个方向,对于环来说,那就更简单了,随便定那个方向都能符合题目的条件.

AC代码:
/* ***********************************************
Author        :xdlove
Created Time  :2015年07月31日 星期五 12时37分29秒
File Name     :a.cpp
 ************************************************ */

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

/**宏定义类
 * **/
#define FOR(i,s,t) for(int i = (s); i < (t); i++)
#define FOR_REV(i,s,t) for(int i = (s - 1); i >= (t); i--)
#define mid ((l + r) >> 1)
#define clr(a) memset(a,0,sizeof(a))
#define lson l,mid,u<<1
#define rson mid+1,r,u<<1|1
#define ls u<<1
#define rs u<<1|1

typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const double pi = acos(-1.0);

/**输入输出挂类模板
 * **/

class Fast
{
    public:
        inline void rd(int &ret)
        {
            char c;
            int sgn;
            if(c = getchar(),c == EOF) return;
            while(c != '-' && (c < '0' || c > '9')) c = getchar();
            sgn = (c == '-') ? -1 : 1;
            ret = (c == '-') ? 0 : (c - '0');
            while(c = getchar(),c >= '0' && c <= '9')
                ret = ret * 10 + c - '0';
            ret *= sgn;
        }

    public:
        inline void pt(int x)
        {
            if(x < 0)
            {
                putchar('-');
                x = -x;
            }
            if(x > 9) pt(x / 10);
            putchar(x % 10 + '0');
        }
};
Fast xd;

const int M = 1e5 + 5;
struct Edge
{
    int to,next,id;
}e[M * 6];
int tot,head[M],pos;
int del[M];

void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v)
{
    e[tot].to = v;
    e[tot].next = head[u];
    e[tot].id = 0;
    head[u] = tot++;
}


bool dfs(int u)
{
    int v,id;
    for(int &i = head[u]; ~i; i = e[i].next)
    {
        v = e[i].to;
        id = e[i].id ^ e[i^1].id;
        if(id) continue;
        e[i].id = 1;
        if(del[v])
        {
            del[v] = 0;
            return true;
        }
        if(dfs(v)) return true;
    }
    return false;
}
            


int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    cin>>T;
    while(T--)
    {
        int n,m;
        scanf("%d %d",&n,&m);
        init();
        FOR(i, 0, m)
        {
            int u,v;
            scanf("%d %d",&u,&v);
            del[u] ^= 1;
            del[v] ^= 1;
            addedge(u,v);
            addedge(v,u);
        }
        for(int i = 1; i <= n; i++)
        {
            if(del[i])
            {
                del[i] = 0;
                dfs(i);
            }
        }
        for(int i = 1; i <= n; i++)
            while(~head[i]) dfs(i);  
        for(int i = 0; i < tot; i += 2)
        {
            if(e[i].id) puts("1");
            else puts("0");
        }
    }
    return 0;
}