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 <bits/stdc++.h>
const int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};
typedef long long ll;
const int N=20;
const ll inf=1ll<<60;
int n,m,g[N][N],ST;
int dp[40000][N],sum[40000];
using namespace std;

/*struct node
{
	int t,next;
}adj[N<<1];
void add(int s,int t)
{
	adj[++now]=(node){t,head[s]};
	head[s]=now;
}*/
inline int read()
{
	char c;int x;
	while (c=getchar(),c<'0' || c>'9');x=c^'0';
	while (c=getchar(),c>='0' && c<='9') x=(x<<1)+(x<<3)+(c^'0');
	return x;
}
void write(int x)
{
	if (x==0) return;
	write(x/10);
	putchar((x%10)+'0');
}
void upd(int& x,int y){if (x==-1) x=y;else x=max(x,y);}
int main()
{
	#ifdef Kay
		freopen ("code.in","r",stdin);
		freopen ("code.out","w",stdout);
	#endif
	cin>>n>>m;
	int i,s,t,w,j,k,l,tot=0;
	for (i=1;i<=m;i++)
	cin>>s>>t>>w,g[s][t]=g[t][s]=w,tot+=w;
	ST=1<<n;
	memset(dp,-1,sizeof(dp));
	for (i=1;i<ST;i++)
	for (s=1;s<=n;s++)
	if ((i>>(s-1))&1)
	for (t=s+1;t<=n;t++)
	if ((i>>(t-1))&1) sum[i]+=g[s][t];
	for (i=0;i<ST;i++)
	if ((i&1)) dp[i][1]=sum[i];
	for (i=1;i<ST;i++)
	for (j=1;j<=n;j++)
	if (dp[i][j]!=-1)
	{
		int x=(i^(ST-1));
		for (k=x;k;k=(k-1)&x)
		for (l=1;l<=n;l++)
		if (((k>>(l-1))&1) && g[j][l]) upd(dp[k|i][l],dp[i][j]+sum[k]+g[j][l]);
	}
	printf ("%d",tot-dp[ST-1][n]);
	return 0;
}