Problem: Mole decided to live in an abandoned mine. The structure of the mine is represented by a simple connected undirected graph which consists of N vertices numbered 1 through N and M edges. The i-th edge connects Vertices a_i and b_i, and it costs c_i yen (the currency of Japan) to remove it.

Mole would like to remove some of the edges so that there is exactly one path from Vertex 1 to Vertex N that does not visit the same vertex more than once. Find the minimum budget needed to achieve this.

Constraints

* 2 \leq N \leq 15
* N-1 \leq M \leq N(N-1)/2
* 1 \leq a_i, b_i \leq N
* 1 \leq c_i \leq 10^{6}
* There are neither multiple edges nor self-loops in the given graph.
* The given graph is connected.

Input

Input is given from Standard Input in the following format:


N M
a_1 b_1 c_1
:
a_M b_M c_M


Output

Print the answer.

Examples

Input

4 6
1 2 100
3 1 100
2 4 100
4 3 100
1 4 100
3 2 100


Output

200


Input

2 1
1 2 1


Output

0


Input

15 22
8 13 33418
14 15 55849
7 10 15207
4 6 64328
6 9 86902
15 7 46978
8 14 53526
1 2 8720
14 12 37748
8 3 61543
6 5 32425
4 11 20932
3 12 55123
8 2 45333
9 12 77796
3 9 71922
12 15 70793
2 4 25485
11 6 1436
2 7 81563
7 11 97843
3 1 40491


Output

133677

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define Rep(i,x) for(int i=head[x];i+1;i=nxt[i])
using namespace std;
int f[20][70005],g[70005];
int con[20][20];
inline int lowbit(int x){return x&(-x);}
int main()
{
	int n,m;scanf("%d%d",&n,&m);int tot=0;
	rep(i,1,m){int x,y,z;scanf("%d%d%d",&x,&y,&z);con[x][y]=con[y][x]=z;tot+=z;}
	memset(f,-1,sizeof(f));
	rep(i,1,(1<<n)-1)rep(j,1,n)if((1<<(j-1))&i)rep(k,j+1,n)if((1<<(k-1))&i)g[i]+=con[j][k];
	rep(i,1,(1<<n)-1)if(i&1)f[1][i]=g[i];
//	rep(i,1,(1<<n)-1)printf("%d ",g[i]);puts("");
	rep(j,0,(1<<n)-1)
	{
		rep(i,1,n)if(f[i][j]>=0)
		{
			int k=((1<<n)-1)^j;
			//printf("Case 0:%d %d\n",k,j);
			for(int p=k;p;p=(p-1)&k)
			{
				//printf("Case 0:%d %d\n",k,p);
				rep(nxt,1,n)if(((1<<(nxt-1))&p) && con[i][nxt])
				{
					f[nxt][j|p]=max(f[nxt][j|p],f[i][j]+con[i][nxt]+g[p]);
				}
			}
		}
	}
	int ans=0;
	rep(i,1,(1<<n)-1)ans=max(ans,f[n][i]);
	printf("%d\n",tot-ans);
	return 0;
}