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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;
 
#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------
 
int N,M;
int mat[101][101];
int dp[1<<15][15];
int cost[1<<15][15];
 
 
void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>M;
	FOR(i,M) {
		cin>>x>>y>>r;
		mat[x-1][y-1]=r;
		mat[y-1][x-1]=r;
	}
	
	FOR(i,1<<N) {
		FOR(x,N) {
			dp[i][x]=1<<30;
			FOR(y,N) if(i&(1<<y)) cost[i][x]+=mat[x][y];
		}
	}
	
	dp[1][0]=0;
	for(int mask=0;mask<1<<N;mask++) {
		FOR(x,N) if((mask&(1<<x)) && dp[mask][x]<1<<30) {
			FOR(y,N) if(mat[x][y] && (mask&(1<<y))==0) {
				int addco=cost[mask][y]-mat[x][y];
				dp[mask|(1<<y)][y]=min(dp[mask|(1<<y)][y],dp[mask][x]+addco);
			}
		}
		FOR(x,N) if((mask&(1<<x)) && dp[mask][x]<1<<30) {
			int nmask=mask^(1<<x);
			int left=((1<<N)-1)^mask;
			for(int i=left; i>0; i=(i-1)&left) {
				int add=0;
				FOR(y,N) if(i&(1<<y)) add+=cost[nmask][y];
				dp[mask|i][x]=min(dp[mask|i][x],dp[mask][x]+add);
			}
		}
	}
	
	
	cout<<dp[(1<<N)-1][N-1]<<endl;
	
}
 
 
int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}