Problem: There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.

The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.

Input

The first line contains a positive integer t (1 ≤ t ≤ 200) — the number of testsets in the input.

Each of the testsets is given in the following way. The first line contains two integers n and m (1 ≤ n ≤ 200, 0 ≤ m ≤ n·(n - 1) / 2) — the number of cities and the number of roads in Berland. 

The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 ≤ u, v ≤ n) — the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.

It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.

Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.

Output

For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.

In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once. 

Example

Input

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


Output

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

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.LinkedList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Wolfgang Beyer
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int t = in.nextInt();
            for (int testCase = 0; testCase < t; testCase++) {
                int n = in.nextInt();
                int m = in.nextInt();
                Graph g = new Graph(n + 1);
                //boolean[][] notFake = new boolean[n][n];
                for (int i = 0; i < m; i++) {
                    int a = in.nextInt();
                    int b = in.nextInt();
                    g.addEdge(a - 1, b - 1);
                    //g.addEdge(in.nextInt() - 1, in.nextInt() - 1);

                    //notFake[a - 1][b - 1] = true;
                    //notFake[b- 1][a - 1] = true;
                }

                //connect all cities with uneven number of edges with eachother
                //now all cities have an even number of edges
                int evenNodes = n;
                //boolean[][] fake = new boolean[n][n];
                int mPlus = m;
                for (int i = 0; i < n; i++) {
                    if (g.getNumberOfEdges(i) % 2 != 0) {
                        evenNodes--;
                        g.addEdge(i, n);
                        //System.out.println("adding additional edge: " + i + " -> " + n);
                        mPlus++;
                    }
                }
                out.println(evenNodes);

            /*List<Integer> path = new ArrayList<>();
            g.getEulerian2(0, new boolean[mPlus], path);
            out.print("path: ");
            for (int i = 0; i < path.size(); i++) out.print(path.get(i) + " ");
            out.println();*/

                boolean[] visited = new boolean[n + 1];

                for (int i = 0; i < n; i++) {
                    if (!visited[i]) {
                        //System.out.println("hier");
                        List<Integer> path = new ArrayList<Integer>();
                        path.add(i);
                        g.getEulerian2(i, new boolean[mPlus], path);
                        //System.out.print("returned: ");
                        for (Object v : path) {
                            //System.out.print(((int) v + 1) + " ");
                            visited[(int) v] = true;
                        }
                        //System.out.println();

                        for (int j = 1; j < path.size(); j++) {
                            if ((path.get(j) < n) && (path.get(j - 1) < n)) {
                                out.println((path.get(j - 1) + 1) + " " + (path.get(j) + 1));
                            }
                        }

                    }
                }

                //System.out.println("even: " + evenNodes);

                //System.out.println("end");
            }

        }

    }

    static class InputReader {
        private static BufferedReader in;
        private static StringTokenizer tok;

        public InputReader(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public String next() {
            try {
                while (tok == null || !tok.hasMoreTokens()) {
                    tok = new StringTokenizer(in.readLine());
                }
            } catch (IOException ex) {
                System.err.println("An IOException was caught :" + ex.getMessage());
            }
            return tok.nextToken();
        }

    }

    static class Graph {
        int numberOfVertices;
        int numberOfEdges;
        LinkedList<int[]>[] adjList;

        public Graph(int vertices) {
            numberOfVertices = vertices;
            numberOfEdges = 0;
            adjList = new LinkedList[numberOfVertices];
            for (int i = 0; i < numberOfVertices; i++) {
                adjList[i] = new LinkedList<int[]>();
                //adjList[i] = new ArrayList<int[]>();
            }
        }

        public void addEdge(int u, int v) {
            //adjList[u].add(v);
            adjList[u].add(new int[]{v, numberOfEdges});
            //adjList[v].add(u);
            adjList[v].add(new int[]{u, numberOfEdges});
            numberOfEdges++;
        }

        public int getNumberOfEdges(int v) {
            return adjList[v].size();
        }

        public void getEulerian2(int current, boolean[] visited, List<Integer> path) {
            for (int[] currentEdge : adjList[current]) {
                if (!visited[currentEdge[1]]) {
                    visited[currentEdge[1]] = true;
                    getEulerian2(currentEdge[0], visited, path);
                    path.add(current);
                    //System.out.println(currentEdge[0] + " -> " + current);
                }
            }
        }

    }
}