Problem Description

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.

Input

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.

Output

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.

Sample Input

Sample Output

  1 /**
  2  * code generated by JHelper
  3  * More info: https://github.com/AlexeyDmitriev/JHelper
  4  * @author xyiyy @https://github.com/xyiyy
  5  */
  6
  7 #include <iostream>
  8 #include <fstream>
  9
 10 //#####################
 11 //Author:fraud
 12 //Blog: http://www.cnblogs.com/fraud/
 13 //#####################
 14 #pragma comment(linker, "/STACK:102400000,102400000")
 15 #include <iostream>
 16 #include <sstream>
 17 #include <ios>
 18 #include <iomanip>
 19 #include <functional>
 20 #include <algorithm>
 21 #include <vector>
 22 #include <string>
 23 #include <list>
 24 #include <queue>
 25 #include <deque>
 26 #include <stack>
 27 #include <set>
 28 #include <map>
 29 #include <cstdio>
 30 #include <cstdlib>
 31 #include <cmath>
 32 #include <cstring>
 33 #include <climits>
 34 #include <cctype>
 35
 36 using namespace std;
 37 #define rep(X, N) for(int X=0;X<N;X++)
 38
 39 const int MAXN = 800010;
 40 int head[MAXN];
 41 int Next[MAXN], To[MAXN];
 42 int vis[MAXN];
 43 int used[100010];
 44 int deg[100010];
 45 int gao;
 46 int tot;
 47
 48 void init(int n) {
 49     tot = 0;
 50     rep(i, n)head[i] = -1;
 51 }
 52
 53 void addedge(int u, int v) {
 54     Next[tot] = head[u];
 55     To[tot] = v;
 56     vis[tot] = 0;
 57     head[u] = tot++;
 58 }
 59
 60 void eular(int u){
 61     used[u] = 1;
 62     int i;
 63     while(head[u]!=-1){
 64     //for(int &i = head[u];i != -1;i = Next[i]){ 65         i = head[u];
 66         head[u] = Next[head[u]];
 67         if(vis[i])continue;
 68         vis[i^1] = 1;
 69         eular(To[i]);
 70     }
 71 }
 72 int Scan() {
 73     int res=0, ch;
 74     while(ch=getchar(), ch<'0'||ch>'9');
 75     res=ch-'0';
 76     while((ch=getchar())>='0'&&ch<='9')
 77         res=res*10+ch-'0';
 78     return res;
 79 }
 80 void Out(int a) {
 81     if (a > 9)
 82         Out(a / 10);
 83     putchar(a % 10 + '0');
 84 }
 85
 86 class hdu5348 {
 87 public:
 88     void solve() {
 89         int t;
 90         t =Scan();//in >> t;
 91         while (t--) {
 92             int n, m;
 93             n = Scan();m=Scan();//in >> n >> m;
 94             init(n);
 95             rep(i, n)deg[i] = 0;
 96             int u, v;
 97             rep(i, m) {
 98                 u = Scan();v= Scan();//in >> u >> v;
 99                 u--, v--;
100                 deg[u]++;
101                 deg[v]++;
102                 addedge(u, v);
103                 addedge(v, u);
104             }
105             gao = -1;
106             rep(i, n) {
107                 if (deg[i] & 1) {
108                     if (gao != -1) {
109                         addedge(i, gao);
110                         addedge(gao, i);
111                         gao = -1;
112                     } else gao = i;
113                 }
114             }
115             rep(i, n) used[i] = 0;
116             /*rep(i,n){
117                 if(!used[i]){
118                     dfs(i);
119                     num++;
120                 }
121             }*/
122             gao = -1;
123             rep(i, n) {
124                 if (!used[i]) {
125                     eular(i);
126                 }
127             }
128             m<<=1;
129             for(int i=1;i<m;i+=2){
130                 if (vis[i])putchar('1');
131                 else putchar('0');
132                 putchar('\n');
133             }
134
135         }
136     }
137 };
138
139
140 int main() {
141     //std::ios::sync_with_stdio(false);
142     //std::cin.tie(0);
143     hdu5348 solver;
144     //std::istream &in(std::cin);
145     //std::ostream &out(std::cout);
146     solver.solve();
147     return 0;
148 }