Problem: Little Chris is participating in a graph cutting contest. He's a pro. The time has come to test his skills to the fullest.

Chris is given a simple undirected connected graph with n vertices (numbered from 1 to n) and m edges. The problem is to cut it into edge-distinct paths of length 2. Formally, Chris has to partition all edges of the graph into pairs in such a way that the edges in a single pair are adjacent and each edge must be contained in exactly one pair.

For example, the figure shows a way Chris can cut a graph. The first sample test contains the description of this graph.

<image>

You are given a chance to compete with Chris. Find a way to cut the given graph or determine that it is impossible!

Input

The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105), the number of vertices and the number of edges in the graph. The next m lines contain the description of the graph's edges. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), the numbers of the vertices connected by the i-th edge. It is guaranteed that the given graph is simple (without self-loops and multi-edges) and connected.

Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.

Output

If it is possible to cut the given graph into edge-distinct paths of length 2, output <image> lines. In the i-th line print three space-separated integers xi, yi and zi, the description of the i-th path. The graph should contain this path, i.e., the graph should contain edges (xi, yi) and (yi, zi). Each edge should appear in exactly one path of length 2. If there are multiple solutions, output any of them.

If it is impossible to cut the given graph, print "No solution" (without quotes).

Examples

Input

8 12
1 2
2 3
3 4
4 1
1 3
2 4
3 5
3 6
5 6
6 7
6 8
7 8


Output

1 2 4
1 3 2
1 4 3
5 3 6
5 6 8
6 7 8

Input

3 3
1 2
2 3
3 1


Output

No solution


Input

3 2
1 2
2 3


Output

1 2 3

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Map.Entry;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.regex.Pattern;

public class K3 {
	static Scanner in = new Scanner();
	static ArrayList<Edge> adj[];
	static int n, m;
	static boolean EdgeVis[];
	static LinkedList<Triple> ans;

	public static void main(String[] args) throws NumberFormatException, IOException {
		/**
		 * Given a simple graph G has |E(G)| = m , if m = 0 (mod 2) <=> G has
		 * paratition into P2 (We can prove that By contradiction) We can solve
		 * this problem by L(G) such that L(G) is line graph but TLE We solve
		 * problem By Good DFS ! O(V+E) linear time Generally , paratition
		 * problem is NP problem :D
		 **/
		n = in.nextInt();
		m = in.nextInt();
		adj = new ArrayList[n];
		EdgeVis = new boolean[m];
		ans = new LinkedList<>();
		for (int i = 0; i < n; i++) {
			adj[i] = new ArrayList<>();
		}
		for (int i = 0; i < m; i++) {
			int u = in.nextInt() - 1;
			int v = in.nextInt() - 1;
			adj[u].add(new Edge(v, i));
			adj[v].add(new Edge(u, i));
		}
		if (m % 2 != 0) {
			System.out.println("No solution");
			System.exit(0);
		}
		dfs(0);
		for (Triple triple : ans) {
			System.out.println(triple);
		}

	}

	static int dfs(int node) {
		ArrayList<Integer> arroud = new ArrayList<>();
		for (Edge e : adj[node]) {
			if (!EdgeVis[e.id]) {
				arroud.add(e.to);
				EdgeVis[e.id] = true;
			}
		}
		Queue<Integer> q = new LinkedList<>();
		for (Integer u : arroud) {
			int ret = dfs(u);
			if (ret == -1) {
				q.add(u);
			} else {
				/*
				 * P2 iso 
				 *    node_____u______ret
				 */
				ans.add(new Triple(node, u, ret));
			}
		}
		while (q.size() >= 2) {
			int poll1 = q.poll();
			int poll2 = q.poll();
			/*
			 * P2 iso
			 *      node
			 *     /   \
			 *    /     \
			 *   poll1   poll2
			 */
			ans.add(new Triple(poll1, node, poll2));
		}
		if (q.size() == 1) {
			return q.poll();
		}
		return -1;
	}

	static class Triple {
		int a, b, c;

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

		@Override
		public String toString() {
			return (a + 1) + " " + (b + 1) + " " + (c + 1);
		}

	}

	static class Edge {
		int to, id;

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

	}

	static class Scanner {// Fast I/O
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String file) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(file));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

	}

}