Codeforces Round #161 (Div. 2)-D. Cycle in Graph

原題鏈接

D. Cycle in Graph

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You've got a undirected graph G, consisting of n nodes. We will consider the nodes of the graph indexed by integers from 1 to n. We know that each node of graph G is connected by edges with at least k other nodes of this graph. Your task is to find in the given graph a simple cycle of length of at least k + 1.

A simple cycle of length d (d > 1) in graph G is a sequence of distinct graph nodes v1, v2, ..., vd such, that nodes v1 and vd are connected by an edge of the graph, also for any integer i (1 ≤ i < d) nodes vi and vi + 1 are connected by an edge of the graph.

Input

The first line contains three integers n, m, k (3 ≤ n, m ≤ 105; 2 ≤ k ≤ n - 1) — the number of the nodes of the graph, the number of the graph's edges and the lower limit on the degree of the graph node. Next m lines contain pairs of integers. The i-th line contains integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the indexes of the graph nodes that are connected by the i-th edge.

It is guaranteed that the given graph doesn't contain any multiple edges or self-loops. It is guaranteed that each node of the graph is connected by the edges with at least k other nodes of the graph.

Output

In the first line print integer r (r ≥ k + 1) — the length of the found cycle. In the next line print r distinct integers v1, v2, ..., vr (1 ≤ vi ≤ n)— the found simple cycle.

It is guaranteed that the answer exists. If there are multiple correct answers, you are allowed to print any of them.

Examples

input

3 3 2
1 2
2 3
3 1

output

3
1 2 3 

input

4 6 3
4 3
1 2
1 3
1 4
2 3
2 4

output

4
3 4 1 2 

從一個點開始深搜，記錄每個節點的深度，若i和j有點相連，且d[i] - d[j] >= k則輸出j到i的路徑

#include <bits/stdc++.h>
#define maxn 100005
#define MOD 1000000007
using namespace std;
typedef long long ll;

vector<int> v[maxn];
int d[maxn], pre[maxn];
int n, m, k;
bool dfs(int j, int f){
	for(int i = 0; i < v[j].size(); i++){
		int h = v[j][i];
		if(h == f)continue;
		if(d[h] == 0){
			d[h] = d[j] + 1;
			pre[h] = j;
			if(dfs(h, j))return true;
		}
		else{
			if(d[h] < d[j] && d[j] - d[h] >= k){
				int s = d[j] - d[h] + 1;
				printf("%d\n", s);
				printf("%d", j);
				s--;
				while(s--){
					printf(" %d", pre[j]);
					j = pre[j];
				}
				return true;
			}
		}
	}
	return false;
}
int main(){
//	freopen("in.txt", "r", stdin);
	int a, b;
	scanf("%d%d%d", &n, &m, &k);
	for(int i = 0; i < m; i++){
		scanf("%d%d", &a, &b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for(int i = 1; i <= n; i++){
		if(d[i] == 0){
			d[i] = 1;
			if(dfs(i, -1))
			 return 0;
		}
	}
}