Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

200

By removing the two edges represented by the red dotted lines in the figure below, the objective can be achieved for a cost of 200 yen.

Sample Input 2

2 1
1 2 1

Sample Output 2

0

It is possible that there is already only one path from Vertex 1 to Vertex N in the beginning.

Sample Input 3

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

Sample Output 3

133677

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

 /*input
4 6
1 2 100
3 1 100
2 4 100
4 3 100
1 4 100
3 2 100
*/
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;
#define IOS ios_base::sync_with_stdio(0); cin.tie(0)
#define ALL(a) a.begin(),a.end()
#define SZ(a) ((int)a.size())
#define F first
#define S second
#define REP(i,n) for(int i=0;i<((int)n);i++)
#define pb push_back
#define MP(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
	out<<'('<<P.F<<','<<P.S<<')';
	return out;
}

//}}}
const int maxn=15;
const int INF=0x3f3f3f3f;

int es[1<<maxn];
int adj[maxn][maxn];
int dp[maxn][1<<maxn];
int main(){
	IOS;
	int n,m;
	cin>>n>>m;
	int meow=1;
	REP(i,n) meow*=2;
	for(int i=1;i<meow;i++){
		int kk=i&-i;
		assert(__lg(kk)<n);
		int tmp=0;
		while(kk%2==0){
			kk/=2;
			tmp++;
		}
	}
	return 0;
}