cf21D Traveling Graph

You are given undirected weighted graph. Find the length of the shortest cycle which starts from the vertex 1 and passes throught all the edges at least once. Graph may contain multiply edges between a pair of vertices and loops (edges from the vertex to itself).

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 15, 0 ≤ m ≤ 2000), n is the amount of vertices, and m is the amount of edges. Following m lines contain edges as a triples x, y, w (1 ≤ x, y ≤ n, 1 ≤ w ≤ 10000), x, y are edge endpoints, and w is the edge length.

Output

Output minimal cycle length or -1 if it doesn't exists.

Example

Input

3 3
1 2 1
2 3 1
3 1 1

Output

3

Input

3 2
1 2 3
2 3 4

Output

14

题意是给一个无向图，要求从1出发回到1，但是必须经过图中每一条边，只上一次

这从不从1出发有什么用呢？这就是个欧拉回路。

然后判一下每个点的度数是不是奇数，如果是奇数就需要再找个奇数度数的点匹配。

最后，匹配直接搜索就行。

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define mkp(a,b) make_pair(a,b)
 #define pi 3.1415926535897932384626433832795028841971
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n,m;
 int v[];
 int fa[];
 int p[],cnt;
 bool mrk[];
 int dis[][];
 LL ans,ans2;
 struct edg{int x,y,z;}e[];
 inline int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
 inline void dfs(int now,LL tot)
 {
     if (now==cnt/+){ans2=min(ans2,tot);return;}
     int x,y;
     for (x=;x<=cnt;x++)if (!mrk[x])break;
     mrk[x]=;
     for (y=x+;y<=cnt;y++)if (!mrk[y]&&dis[p[x]][p[y]]<=)
     {
         mrk[y]=;
         dfs(now+,tot+dis[p[x]][p[y]]);
         mrk[y]=;
     }
     mrk[x]=;
 }
 int main()
 {
     n=read();m=read();
     for (int i=;i<=n;i++)fa[i]=i;
     for (int i=;i<n;i++)
     for (int j=i+;j<=n;j++)
         dis[i][j]=dis[j][i]=<<;
     for (int i=;i<=m;i++)
     {
         e[i].x=read();
         e[i].y=read();
         e[i].z=read();v[e[i].x]++;v[e[i].y]++;
         if (e[i].z<dis[e[i].x][e[i].y])dis[e[i].x][e[i].y]=dis[e[i].y][e[i].x]=e[i].z;
         ans+=e[i].z;
         int xx=getfa(e[i].x),yy=getfa(e[i].y);
         if (xx>yy)swap(xx,yy);
         fa[xx]=yy;
     }
     for (int i=;i<=m;i++)
     {
         if (getfa(e[i].x)==getfa()&&getfa(e[i].y)==getfa())continue;
         puts("-1");
         return ;
     }
     for (int k=;k<=n;k++)
         for (int i=;i<=n;i++)
             for (int j=;j<=n;j++)
                 if (dis[i][j]>dis[i][k]+dis[k][j])
                     dis[i][j]=dis[i][k]+dis[k][j];
     for (int i=;i<=n;i++)if (v[i]&)p[++cnt]=i;
     if (cnt&){puts("-1");return ;}
     ans2=1ll<<;
     dfs(,ans);
     printf("%lld\n",ans2);
 }

cf21D

You are given undirected weighted graph. Find the length of the shortest cycle which starts from the vertex 1 and passes throught all the edges at least once. Graph may contain multiply edges between a pair of vertices and loops (edges from the vertex to itself).

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 15, 0 ≤ m ≤ 2000), n is the amount of vertices, and m is the amount of edges. Following m lines contain edges as a triples x, y, w (1 ≤ x, y ≤ n, 1 ≤ w ≤ 10000), x, y are edge endpoints, and w is the edge length.

Output

Output minimal cycle length or -1 if it doesn't exists.

Example

Input

3 3
1 2 1
2 3 1
3 1 1

Output

3

Input

3 2
1 2 3
2 3 4

Output

14