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 <stdio.h>
#include <algorithm>
#include <iostream>
#include <memory.h>
#include <queue>

using namespace std;
typedef long long LL;
const int maxn = 20;

int e[maxn][maxn],n,m,bin[1<<15];
LL dp[1<<15][maxn];
LL cs[1<<15][maxn]; 
LL cst[1<<15],sum;
void upd(LL &x,LL y) {if (x<y) x=y;}
int main()
{
	#ifdef LOCAL_JUDGE
		freopen("mine.in","r",stdin);
		freopen("mine.out","w",stdout);
	#endif
	scanf("%d %d",&n,&m);
	for (int i=0;i<n;i++) bin[1<<i]=i;
	for (int u,v,w,i=1;i<=m;i++)
		scanf("%d %d %d",&u,&v,&w),
		u--,v--,e[u][v]=e[v][u]=w,sum+=w;
	for (int i=0;i<n;i++)
	for (int j=1;j<(1<<n);j++)
		cs[j][i]=cs[j^(j&-j)][i]+e[bin[j&-j]][i];
	for (int i=1;i<(1<<n);i++)
		cst[i]=cst[i^(i&-i)]+cs[i^(i&-i)][bin[i&-i]];
	memset(dp,-1,sizeof dp);
	dp[1][0]=0;
	for (int s=0;s<(1<<n);s++)
	for (int u=0;u<n;u++)
	if ((s>>u&1)&&dp[s][u]!=-1) {
		for (int t=0;t<n;t++)
		if (!(s>>t&1)&&e[u][t])
			upd(dp[s^(1<<t)][t],e[u][t]+dp[s][u]);
		int o=((1<<n)-1)^s;
		for (int t=o;t;t=t-1&o)
			upd(dp[s^t][u],dp[s][u]+cst[t]+cs[t][u]);
	}
	printf("%lld",sum-dp[(1<<n)-1][n-1]);
	return 0;
}