Description

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.

<center> </center>

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 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).

Sample Input

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

题意：

给n个点，m条边，将这个图拆成多个子图，每个子图里面有两条相邻的边（也就是每个子图 3个点2条边），当然每条边只能属于一个子图，如果不能拆分成功，则No solution， 否则，将每个子图的3个顶点（x，y，z）输出 （这个子图包含 x到y 的边 ， y到z的边 ）

首先：m是奇数肯定不行；偶数时，随便找一个点，跟他连接的点的个数，两两和这个点连，如果剩了，当前的点，剩的点，剩的点的儿子组成一个子图。其实挺简单的题，非要纠结在点怎么分配，点的分配还不是得跟着边嘛，只能遍历一次的是边，又不是点……

#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
struct edge
{
    int id,num;
}t;
vector<edge>v[100005];
bool flag[100005];
int dfs(int x)
{
    queue<int>q;
   // cout<<"bbbb"<<endl;
    edge k;
    int l=v[x].size();
    for(int i=0;i<l;i++)
    {
        k=v[x][i];
        if(flag[k.id]) continue;
        flag[k.id]=1;
        int r=dfs(k.num);
        if(r) printf("%d %d %d\n",x,k.num,r);
        else q.push(k.num);
    }
  //  cout<<"ccc"<<endl;
    while(q.size()>=2)
    {
        int a=q.front();
        q.pop();
        int b=q.front();
        q.pop();
        printf("%d %d %d\n",a,x,b);
    }
    if(!q.empty())
    {
        return q.front();
    }
    return 0;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        int a,b;
        for(int i=0;i<=n;i++) v[i].clear();
        memset(flag,0,sizeof(flag));
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            t.id=i;
            t.num=b;
            v[a].push_back(t);
            t.num=a;
            v[b].push_back(t);
          //  cout<<"aaaa"<<endl;
        }
        if(m&1)printf("No solution\n");
        else dfs(1);
    }
    return 0;
}