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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define MOD 1000000007LL
#define INF 1000000000LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

struct edge{
	int t,c;
	edge(){}
	edge(int tt,int cc){
		t=tt;
		c=cc;
	}
};

vector<edge> G[20];

int n,m;
int a[500];
int b[500];
int c[500];
int dp[1<<15][15];
int cost[1<<15];
int dp2[1<<15][15];
int node[15];
int node2[15];

int main(void){
	int s=n,t=s+1;
	scanf("%d%d",&n,&m);
	int sum=0;
	for(int i=0;i<m;i++){
		scanf("%d%d%d",&a[i],&b[i],&c[i]);
		a[i]--;
		b[i]--;
		sum+=c[i];
		G[a[i]].push_back(edge(b[i],c[i]));
		G[b[i]].push_back(edge(a[i],c[i]));
	}
	for(int i=0;i<(1<<n);i++){
		for(int j=0;j<n;j++){
			if(!(i>>j & 1))continue;
			for(int k=0;k<G[j].size();k++){
				int nv=G[j][k].t;
				if(nv>j && ((i>>nv) & 1)){
					cost[i]+=G[j][k].c;
				}
			}
		}
	}
	memset(dp,-1,sizeof(dp));
	dp[1][0]=0;
	for(int i=0;i<(1<<n);i++){
		for(int j=0;j<n;j++){
			if(dp[i][j]>=0){
				for(int k=0;k<G[j].size();k++){
					int nv=G[j][k].t;
					int nc=G[j][k].c;
					if(!(i>>nv & 1)){
						int nbit=i+(1<<nv);
						dp[i+(1<<nv)][nv]=max(dp[i+(1<<nv)][nv],dp[i][j]+nc);
					}
				}
			}
		}
	}
	int res=0;
	for(int i=0;i<(1<<n);i++){
		if(!(i>>0 & 1))continue;
		if(!(i>>(n-1) & 1))continue;
		if(dp[i][n-1]==-1)continue;
		int cnt=0;
		int cnt2=0;
		for(int j=0;j<n;j++){
			if(i>>j & 1){
				node[cnt++]=j;
			}else{
				node2[cnt2++]=j;
			}
		}
		for(int j=0;j<(1<<cnt2);j++){
			for(int k=0;k<=cnt;k++){
				dp2[j][k]=-1;
			}
		}
		dp2[0][0]=0;
		for(int j=0;j<cnt;j++){
			for(int k=0;k<(1<<cnt2);k++){
				if(dp2[k][j]>=0){
					for(int l=k;l<(1<<cnt2);l=((l+1)|k)){
						int rest=l-k;
						int bit=1<<node[j];
						for(int lb=0;lb<cnt2;lb++){
							if(rest>>lb & 1){
								bit+=1<<(node2[lb]);
							}
						}
						dp2[l][j+1]=max(dp2[l][j+1],dp2[k][j]+cost[bit]);
					}
				}
			}
		}
		if(dp[i][n-1]>=0 && dp2[(1<<cnt2)-1][cnt]>=0)res=max(res,dp[i][n-1]+dp2[(1<<cnt2)-1][cnt]);
	}
	printf("%d\n",sum-res);
	return 0;
}