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;
const int MAXN = 15 + 5;
const int ALL = (1<<15) + 5;
const int inf = 0x3f3f3f3f;

inline void chk_max(int &a,int b){ if(a<b) a=b;}

int lb[ALL];
#define bit(x,i) (((x)>>(i))&1)
#define lowbit(x) ((x)&-(x))

int g[MAXN][MAXN];
int cost[ALL],f[ALL][MAXN];

int main(void)
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1; i<=m; ++i)
	{
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		g[u][v]=g[v][u]=w;
	}
	
	int all = (1<<n)-1;
	
	for(int mask=0; mask<=all; ++mask)
	{
		for(int i=1; i<=n; ++i) if(bit(mask,i-1))
			for(int j=1; j<i; ++j) if(bit(mask,j-1))
				cost[mask]+=g[i][j];
	}
	
	for(int mask=0; mask<=all; ++mask)
	{
		for(int i=1; i<=n; ++i) f[mask][i]=-inf;
		if(mask&1) f[mask][1]=cost[mask];
	}
	int ans=0;
	for(int mask=1; mask<=all; mask+=2)
	{
		if(bit(mask,n-1))
		{
			chk_max(ans,f[mask][n]);
			continue;
		}
		for(int i=1; i<=n; ++i) if(bit(mask,i-1))
		{
			if(f[mask][i]==-inf) continue;
			for(int mask2=all^mask; mask2; mask2=(mask2-1)&(all^mask))
			{
				for(int j=1; j<=n; ++j) if(bit(mask2,j-1))
					if(g[i][j]>0)
						chk_max(f[mask|mask2][j], f[mask][i]+g[i][j]+cost[mask2]);
			}
		}
	}
	
	printf("%d",cost[all]-ans);
	return 0;
}