Problem: All Berland residents are waiting for an unprecedented tour of wizard in his Blue Helicopter over the cities of Berland!

It is well-known that there are n cities in Berland, some pairs of which are connected by bidirectional roads. Each pair of cities is connected by no more than one road. It is not guaranteed that the road network is connected, i.e. it is possible that you can't reach some city from some other.

The tour will contain several episodes. In each of the episodes:

  * the wizard will disembark at some city x from the Helicopter; 
  * he will give a performance and show a movie for free at the city x; 
  * he will drive to some neighboring city y using a road; 
  * he will give a performance and show a movie for free at the city y; 
  * he will drive to some neighboring to y city z; 
  * he will give a performance and show a movie for free at the city z; 
  * he will embark the Helicopter and fly away from the city z. 



It is known that the wizard doesn't like to use roads, so he agrees to use each road at most once (regardless of direction). In other words, for road between a and b he only can drive once from a to b, or drive once from b to a, or do not use this road at all.

The wizards wants to plan as many episodes as possible without violation the above rules. Help the wizard!

Please note that the wizard can visit the same city multiple times, the restriction is on roads only.

Input

The first line contains two integers n, m (1 ≤ n ≤ 2·105, 0 ≤ m ≤ 2·105) — the number of cities and the number of roads in Berland, respectively.

The roads description follow, one in each line. Each description is a pair of two integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the ids of the cities connected by the i-th road. It is guaranteed that there are no two roads connecting the same pair of cities. Every road is bidirectional. The cities are numbered from 1 to n.

It is possible that the road network in Berland is not connected.

Output

In the first line print w — the maximum possible number of episodes. The next w lines should contain the episodes in format x, y, z — the three integers denoting the ids of the cities in the order of the wizard's visits.

Examples

Input

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


Output

2
1 4 2
4 3 2


Input

5 8
5 3
1 2
4 5
5 1
2 5
4 3
1 4
3 2


Output

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

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.Stream;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.InputStream;

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

    static class TaskD {
        public List<TaskD.Edge>[] graph;
        public List<TaskD.Path> res;
        public boolean[] vis;
        public boolean[] pp;

        public void solve(int testNumber, InputReader in, OutputWriter out) {
            int n = in.nextInt(), m = in.nextInt();
            graph = LUtils.genArrayList(n);
            for (int i = 0; i < m; i++) {
                int a = in.nextInt() - 1, b = in.nextInt() - 1;
                graph[a].add(new TaskD.Edge(b, i));
                graph[b].add(new TaskD.Edge(a, i));
            }

            res = new ArrayList<>();
            vis = new boolean[n];
            pp = new boolean[n];
            for (int i = 0; i < n; i++) {
                if (!vis[i]) {
                    dfs(i);
                }
            }

            out.println(res.size());
            for (TaskD.Path p : res) {
                out.printf("%d %d %d\n", p.a + 1, p.b + 1, p.c + 1);
            }
        }

        public int dfs(int node) {
            if (vis[node]) return -1;
            pp[node] = true;
            vis[node] = true;
            int unmatched = -1;
            for (TaskD.Edge e : graph[node]) {
                if (pp[e.to]) continue;
                int x = dfs(e.to);
                if (x != -1) {
                    res.add(new TaskD.Path(node, e.to, x));
                } else if (unmatched != -1) {
                    res.add(new TaskD.Path(unmatched, node, e.to));
                    unmatched = -1;
                } else {
                    unmatched = e.to;
                }
            }
            pp[node] = false;
            return unmatched;
        }

        static class Edge {
            public int to;
            public int idx;

            public Edge(int to, int idx) {
                this.to = to;
                this.idx = idx;
            }

        }

        static class Path {
            public int a;
            public int b;
            public int c;

            public Path(int a, int b, int c) {
                this.a = a;
                this.b = b;
                this.c = c;
            }

        }

    }

    static class LUtils {
        public static <E> List<E>[] genArrayList(int size) {
            return Stream.generate(ArrayList::new).limit(size).toArray(List[]::new);
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (this.numChars == -1) {
                throw new InputMismatchException();
            } else {
                if (this.curChar >= this.numChars) {
                    this.curChar = 0;

                    try {
                        this.numChars = this.stream.read(this.buf);
                    } catch (IOException var2) {
                        throw new InputMismatchException();
                    }

                    if (this.numChars <= 0) {
                        return -1;
                    }
                }

                return this.buf[this.curChar++];
            }
        }

        public int nextInt() {
            int c;
            for (c = this.read(); isSpaceChar(c); c = this.read()) {
                ;
            }

            byte sgn = 1;
            if (c == 45) {
                sgn = -1;
                c = this.read();
            }

            int res = 0;

            while (c >= 48 && c <= 57) {
                res *= 10;
                res += c - 48;
                c = this.read();
                if (isSpaceChar(c)) {
                    return res * sgn;
                }
            }

            throw new InputMismatchException();
        }

        public static boolean isSpaceChar(int c) {
            return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
        }

    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void printf(String format, Object... objects) {
            writer.printf(format, objects);
        }

        public void close() {
            writer.close();
        }

        public void println(int i) {
            writer.println(i);
        }

    }
}