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<algorithm>
using namespace std;
#define N 17
int v[N][N],n,m,a,b,c,as;
int dp[67890][N],su[67890],s1;
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)scanf("%d%d%d",&a,&b,&c),v[a][b]=v[b][a]=c,s1+=c;
	for(int i=1;i<1<<n;i++)
	for(int j=1;j<=n;j++)
	for(int k=j+1;k<=n;k++)
	if((i&(1<<j-1))&&(i&(1<<k-1)))su[i]+=v[j][k];
	for(int i=0;i<1<<n;i++)for(int j=1;j<=n;j++)dp[i][j]=-1e9;
	for(int i=1;i<1<<n;i+=2)dp[i][1]=su[i];
	for(int i=1;i<1<<n;i++)
	for(int j=1;j<=n;j++)
	if(i&(1<<j-1))
	for(int k=((1<<n)-1)^i;k;k=(k-1)&(((1<<n)-1)^i))
	for(int l=1;l<=n;l++)
	if(k&(1<<l-1))
	if(v[j][l])
	dp[i|k][l]=max(dp[i|k][l],dp[i][j]+su[k]+v[j][l]);
	int as=0;
	for(int i=1;i<1<<n;i++)as=max(as,dp[i][n]);
	printf("%d\n",s1-as);
}