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.PrintWriter;
import java.util.HashSet;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.io.InputStream;

/**
 * @author khokharnikunj8
 */

public class Main {
    public static void main(String[] args) {
        new Thread(null, new Runnable() {
            public void run() {
                new Main().solve();
            }
        }, "1", 1 << 26).start();
    }

    void solve() {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        ScanReader in = new ScanReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DWizardsTour solver = new DWizardsTour();
        solver.solve(1, in, out);
        out.close();
    }

    static class DWizardsTour {
        int[][] G;
        boolean[] visited;
        int[][] ans;
        int tl;
        HashSet<Long> set;

        public void dfs(int n, int p) {
            visited[n] = true;
            int last_node = -1;
            for (int i : G[n]) if (!visited[i]) dfs(i, n);
            for (int i : G[n]) {
                if (i == p) continue;
                if (!set.contains(1000000l * n + i)) {
                    if (last_node != -1) {
                        ans[tl][0] = last_node;
                        ans[tl][1] = n;
                        ans[tl][2] = i;
                        tl++;
                        set.add(1000000l * n + last_node);
                        set.add(1000000l * last_node + n);
                        set.add(1000000l * n + i);
                        set.add(1000000l * i + n);
                        last_node = -1;
                    } else last_node = i;

                }
            }
            if (last_node != -1 && p != -1) {
                ans[tl][0] = last_node;
                ans[tl][1] = n;
                ans[tl][2] = p;
                tl++;
                set.add(1000000l * n + last_node);
                set.add(1000000l * last_node + n);
                set.add(1000000l * n + p);
                set.add(1000000l * p + n);
            }


        }

        public void solve(int testNumber, ScanReader in, PrintWriter out) {
            tl = 0;
            int n = in.scanInt();
            int m = in.scanInt();
            ans = new int[2 * m + 1][3];
            set = new HashSet<>();
            int[] from = new int[m];
            int[] to = new int[m];
            visited = new boolean[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = in.scanInt();
                to[i] = in.scanInt();
            }
            G = CodeHash.packGraph(from, to, n);
            for (int i = 1; i <= n; i++) if (!visited[i]) dfs(i, -1);
            out.println(tl);
            for (int i = 0; i < tl; i++) out.println(ans[i][0] + " " + ans[i][1] + " " + ans[i][2]);


        }

    }

    static class CodeHash {
        public static int[][] packGraph(int[] from, int[] to, int n) {
            int[][] g = new int[n + 1][];
            int p[] = new int[n + 1];
            for (int i : from) p[i]++;
            for (int i : to) p[i]++;
            for (int i = 0; i <= n; i++) g[i] = new int[p[i]];
            for (int i = 0; i < from.length; i++) {
                g[from[i]][--p[from[i]]] = to[i];
                g[to[i]][--p[to[i]]] = from[i];
            }
            return g;
        }

    }

    static class ScanReader {
        private byte[] buf = new byte[4 * 1024];
        private int index;
        private BufferedInputStream in;
        private int total;

        public ScanReader(InputStream inputStream) {
            in = new BufferedInputStream(inputStream);
        }

        private int scan() {
            if (index >= total) {
                index = 0;
                try {
                    total = in.read(buf);
                } catch (Exception e) {
                    e.printStackTrace();
                }
                if (total <= 0) return -1;
            }
            return buf[index++];
        }

        public int scanInt() {
            int integer = 0;
            int n = scan();
            while (isWhiteSpace(n)) n = scan();
            int neg = 1;
            if (n == '-') {
                neg = -1;
                n = scan();
            }
            while (!isWhiteSpace(n)) {
                if (n >= '0' && n <= '9') {
                    integer *= 10;
                    integer += n - '0';
                    n = scan();
                }
            }
            return neg * integer;
        }

        private boolean isWhiteSpace(int n) {
            if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
            else return false;
        }

    }
}