2020 BIT冬训-图&&DFS&&BFS P - Underground Lab CodeForces - 780E

Problem Description

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  vertices before the lab explodes.

Your task is to choose starting vertices and searching routes for the clones. Each route can have at most  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 x_i and y_i (1 ≤ x_i, y_i ≤ 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 c_i () — the number of vertices visited by i-th clone, followed by c_i 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.

 

由于这是一个无向图，我们可以发现dfs+回溯遍历一个n个顶点的图最多只需要走2*n步。（例如以1为中心其他每个点都与她相连）

故dfs+回溯将所有步数存在一个数组当中。

然后按k个点每个点走最大步数的分配。

如果k个点有剩余（即一步都分配不到，则输出1 1即可）

AC代码如下：

#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <queue>
#define INF 0x3f3f3f3f
#define MAXN 200100
#define N 1111
#define eps 1e-6
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
vector<int> g[MAXN];
bool vis[MAXN];
int n,m,k,path[MAXN*2],cnt;
void dfs(int from){
    vis[from] = 1;
    path[++cnt] = from;
    for (int i = 0; i < (int)g[from].size(); i++){
        int to = g[from][i];
        if (!vis[to]){
            dfs(to);
            path[++cnt]=from;//将回溯的点记录 
        }
    }
}
int main(){
    while (~scanf("%d%d%d",&n,&m,&k)){
        cnt=0;
        memset(vis,0,sizeof(vis));
        for (int i=1; i<=n;i++)
            g[i].clear();
        for (int i=0; i<m;i++){
            int u, v;
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs(1);
        int maxC=(2*n+k-1)/k;//maxC为该点最多能走几步 
        for (int i=0;i<k;i++){
            int c = min(maxC,cnt);
            if(c){
                printf("%d",c);
                for (int j=0;j<c&&cnt;j++)
                    printf(" %d",path[cnt--]);
                printf("\n");
            }
            else
                printf("1 1\n");
        }
    }
    return 0;
}