Problem: The evil Bumbershoot corporation produces clones for gruesome experiments in a vast underground lab. On one occasion, the corp cloned a boy Andryusha who was smarter than his comrades. Immediately Andryusha understood that something fishy was going on there. He rallied fellow clones to go on a feud against the evil corp, and they set out to find an exit from the lab. The corp had to reduce to destroy the lab complex.

The lab can be pictured as a connected graph with n vertices and m edges. k clones of Andryusha start looking for an exit in some of the vertices. Each clone can traverse any edge once per second. Any number of clones are allowed to be at any vertex simultaneously. Each clone is allowed to stop looking at any time moment, but he must look at his starting vertex at least. The exit can be located at any vertex of the lab, hence each vertex must be visited by at least one clone.

Each clone can visit at most <image> vertices before the lab explodes.

Your task is to choose starting vertices and searching routes for the clones. Each route can have at most <image> vertices.

Input

The first line contains three integers n, m, and k (1 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105, 1 ≤ k ≤ n) — the number of vertices and edges in the lab, and the number of clones.

Each of the next m lines contains two integers xi and yi (1 ≤ xi, yi ≤ n) — indices of vertices connected by the respective edge. The graph is allowed to have self-loops and multiple edges.

The graph is guaranteed to be connected.

Output

You should print k lines. i-th of these lines must start with an integer ci (<image>) — the number of vertices visited by i-th clone, followed by ci integers — indices of vertices visited by this clone in the order of visiting. You have to print each vertex every time it is visited, regardless if it was visited earlier or not.

It is guaranteed that a valid answer exists.

Examples

Input

3 2 1
2 1
3 1


Output

3 2 1 3


Input

5 4 2
1 2
1 3
1 4
1 5


Output

3 2 1 3
3 4 1 5

Note

In the first sample case there is only one clone who may visit vertices in order (2, 1, 3), which fits the constraint of 6 vertices per clone.

In the second sample case the two clones can visited vertices in order (2, 1, 3) and (4, 1, 5), which fits the constraint of 5 vertices per clone.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.Math;
import java.math.BigInteger;

public class Problem {

	public static void main(String[] args) throws IOException {

		new Problem().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer tok;
	static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
	Problem() throws FileNotFoundException{

		if (ONLINE_JUDGE) {
			in = new BufferedReader(new InputStreamReader(System.in)); 
            out = new PrintWriter(System.out); 
		} else {
			in = new BufferedReader(new FileReader("input.txt"));
			out = new PrintWriter("output.txt");
		}
		tok = new StringTokenizer("");
	}

	// ======================================================

	int INF = Integer.MAX_VALUE;
	List<Integer>[] graph;
	boolean[] used;
	long[] time;
	int[] def;
	int[] ans;
	ArrayList<Integer> ansPath = new ArrayList<>();
	boolean flag = false;
	int maxLength;
	int curClone = 0;
	void solve() throws IOException {
		int n = rI();
		int m = rI();
		int k = rI();
		maxLength = (2*n)/k;
		if(2*n%k>0){
			maxLength++;
		}
		ans = new int[n];
		used = new boolean[n];
		graph = new ArrayList[n];
		for(int i=0;i<n;++i){
			graph[i] = new ArrayList<>();
		}
		for(int i=0;i<m;++i){
			int from = rI()-1;
			int to = rI()-1;
			graph[from].add(to);
			graph[to].add(from);
		}
		dfs(0);
		List<Integer>[] ways = new ArrayList[k];
		for(int i=0;i<k;++i){
			ways[i] = new ArrayList<>();
		}
		for(int i=0;i<ansPath.size();++i){
			ways[i/maxLength].add(ansPath.get(i));
		}
		for(int i=0;i<k;++i){
			if(ways[i].isEmpty()){
				ways[i].add(1);
			}
		}
		for(int i=0;i<k;++i){
			out.print(ways[i].size()+" ");
			for(int ans: ways[i]){
				out.print(ans+" ");
			}
			out.println();
		}
	}
	void dfs(int from){
		used[from] = true;
		boolean flag = false;
		ansPath.add(from+1);
		for(int to: graph[from]){
			if(!used[to]){
				flag = true;
				dfs(to);
				ansPath.add(from+1);
			}
		}
		
	}
	// ======================================================
	void run() throws IOException {
		solve();
		out.close();
	}


	int sqrt(double m){
		int l = 0;
		int r = 1000000000+9;
		int i=1000;
		while(r-l>1){
			int mid = (r+l)/2;
			if(mid*mid>m){
				r = mid;
			}else{
				l = mid;
			}
		}
		return l;
	}
	int countPow(int m,int n){
		int ans = 0;
		while(m%n==0&&m>0){
			ans++;
			m/=n;
		}
		return ans;
	}
	long binPow(long a, long b){
		if(b == 0){
			return 1;
		}
		if(b%2==1){
			return a*binPow(a,b-1);
		}else{
			long c = binPow(a,b/2);
			return c*c;
		}
		
	}
	long gcd(long a, long b){
		return b==0?a:gcd(b,a%b);
	}
	
	long pow(long x, long k){
		long ans=1;
		for(int i=0;i<k;++i){
			ans*=x;
		}
		return ans;
	}
	//////////////////////////////////////////////////////////////////////


	String delimiter = " ";
	String readLine() throws IOException {
		return in.readLine();
	}
	String rS() throws IOException {
		while (!tok.hasMoreTokens()) {
			String nextLine = readLine();
			if (null == nextLine)
				return null;

			tok = new StringTokenizer(nextLine);
		}

		return tok.nextToken(delimiter);
	}
	int rI() throws IOException {
		return Integer.parseInt(rS());
	}

	long rL() throws IOException {
		return Long.parseLong(rS());
	}
	int[] rA(int b) {
		int a[] = new int[b];
		for (int i = 0; i < b; i++) {
			try {
				a[i] = rI();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return a;
	}
	
	int[] readSortedIntArray(int size) throws IOException {
	        Integer[] array = new Integer[size];

	        for (int index = 0; index < size; ++index) {
	            array[index] = rI();
	        }
	        Arrays.sort(array);

	        int[] sortedArray = new int[size];
	        for (int index = 0; index < size; ++index) {
	            sortedArray[index] = array[index];
	        }

	        return sortedArray;
	    }

}