Problem: You are given an undirected unweighted connected graph consisting of n vertices and m edges. It is guaranteed that there are no self-loops or multiple edges in the given graph.

Your task is to find any spanning tree of this graph such that the degree of the first vertex (vertex with label 1 on it) is equal to D (or say that there are no such spanning trees). Recall that the degree of a vertex is the number of edges incident to it.

Input

The first line contains three integers n, m and D (2 ≤ n ≤ 2 ⋅ 10^5, n - 1 ≤ m ≤ min(2 ⋅ 10^5, (n(n-1))/(2)), 1 ≤ D < n) — the number of vertices, the number of edges and required degree of the first vertex, respectively.

The following m lines denote edges: edge i is represented by a pair of integers v_i, u_i (1 ≤ v_i, u_i ≤ n, u_i ≠ v_i), which are the indices of vertices connected by the edge. There are no loops or multiple edges in the given graph, i. e. for each pair (v_i, u_i) there are no other pairs (v_i, u_i) or (u_i, v_i) in the list of edges, and for each pair (v_i, u_i) the condition v_i ≠ u_i is satisfied.

Output

If there is no spanning tree satisfying the condition from the problem statement, print "NO" in the first line.

Otherwise print "YES" in the first line and then print n-1 lines describing the edges of a spanning tree such that the degree of the first vertex (vertex with label 1 on it) is equal to D. Make sure that the edges of the printed spanning tree form some subset of the input edges (order doesn't matter and edge (v, u) is considered the same as the edge (u, v)).

If there are multiple possible answers, print any of them.

Examples

Input


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


Output


YES
2 1
2 3
3 4


Input


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


Output


YES
1 2
1 3
4 1


Input


4 4 3
1 2
1 4
2 3
3 4


Output


NO

Note

The picture corresponding to the first and second examples: <image>

The picture corresponding to the third example: <image>

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class spanningTreeWithOneFixedDegree {
	static int n, m, D;
	static ArrayDeque[] edges, subtrees;
	static int[] u, v, map;
	static boolean[] inTree, seen;
	public static void main(String[] args) {
		FS sc = new FS();
		PrintWriter pw = new PrintWriter(System.out);
		n = sc.nextInt();
		m = sc.nextInt();
		D = sc.nextInt();
		edges = new ArrayDeque[n];
		for(int i = 0; i < n; ++i) edges[i] = new ArrayDeque<>();
		u = new int[m];
		v = new int[m];
		for(int i = 0; i < m; ++i) {
			u[i] = sc.nextInt() - 1;
			v[i] = sc.nextInt() - 1;
			edges[u[i]].add(new edge(v[i], i));
			edges[v[i]].add(new edge(u[i], i));
		}
		
		//keeps track of what edges are currently in our tree
		inTree = new boolean[m];
		seen = new boolean[n];
		seen[0] = true;
		//current degree of root node
		int currD = edges[0].size();
		//potentially break early
		if(currD < D) {
			System.out.println("NO");
			return;
		}
		//tells you what subtree this node is currently in; subtrees will be indexed by the index of the first node in the subtree (the parent)
		map = new int[n];
		for(edge e : (ArrayDeque<edge>) edges[0]) {
			inTree[e.i] = true;
			seen[e.v] = true;
		}
		
		subtrees = new ArrayDeque[n];
		for(int i = 0; i < n; ++i) subtrees[i] = new ArrayDeque<>();
		//do the dfs on all adjacent nodes to root to get their sizes and also to set inTree accordingly
		for(edge e : (ArrayDeque<edge>) edges[0]) dfs(e.v, e.v);
		
		//now loop over all edges that aren't already in our tree and see if they can union two subtrees to bring down currD
		//too lazy to make a map for every edge that connects to root so we're just gonna say if we delete an edge from root to i, then notToRoot[i] = true
		boolean[] notToRoot = new boolean[n];
		for(int i = 0; i < m; ++i) {
			if(currD == D) break;
			
			if(inTree[i] || u[i] == 0 || v[i] == 0) continue;
			if(map[u[i]] != map[v[i]]) {
				//greedily union these two subtrees
				inTree[i] = true;
				int a = map[u[i]], b = map[v[i]];
				if(subtrees[a].size() < subtrees[b].size()) {
					int temp = a;
					a = b;
					b = temp;
				}
				
				//delete edge from root to b
				notToRoot[b] = true;
				
				//now we merge everything in b to a
				for(int curr : (ArrayDeque<Integer>) subtrees[b]) {
					map[curr] = a;
					subtrees[a].add(curr);
				}
				
				--currD;
			}
		}
		
		if(currD > D) {
			System.out.println("NO");
			return;
		}
		
		pw.println("YES");
		for(int i = 0; i < m; ++i) {
			int a = u[i], b = v[i];
			if(inTree[i] && !(a == 0 && notToRoot[b]) && !(b == 0 && notToRoot[a])) {
				pw.println((a + 1) + " " + (b + 1));
			}
		}
		pw.flush();
	}
	static void dfs(int idx, int hyperPar) {
		seen[idx] = true;
		map[idx] = hyperPar;
		subtrees[hyperPar].add(idx);
		
		for(edge e : (ArrayDeque<edge>) edges[idx]) {
			if(!seen[e.v]) {
				inTree[e.i] = true; 
				dfs(e.v, hyperPar);
			}
		}
	}
	static class edge{
		int v, i;
		edge(int vv, int ii){
			v = vv;
			i = ii;
		}
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}