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 <bits/stdc++.h>
using namespace std;
typedef long long lint;
const int MAXN = 15;

struct edg{
	int s, e, x;
};

vector<edg> v;
int cyc[1 << MAXN];
int dp[1 << MAXN][MAXN];
int adj[MAXN][MAXN];
int n, m;

int main(){
	scanf("%d %d",&n,&m);
	for(int i=0; i<MAXN; i++){
		for(int j=0; j<MAXN; j++){
			adj[i][j] = -1e9;
		}
	}
	int ans = 0;
	for(int i=0; i<m; i++){
		int s, e, x;
		scanf("%d %d %d",&s,&e,&x);
		adj[s-1][e-1] = x;
		adj[e-1][s-1] = x;
		v.push_back({s-1, e-1, x});
	}
	for(int i=0; i<(1<<n); i++){
		for(auto &j : v){
			if((i >> j.s) % 2 == 1 & (i >> j.e) % 2 == 1){
				cyc[i] += j.x;
			}
		}
	}
	for(int i=0; i<(1<<n); i++){
		for(int j=0; j<n; j++){
			dp[i][j] = -1e9;
			if((i >> j) % 2 == 0){
				continue;
			}
			if(j == 0){
				dp[i][j] = cyc[i];
			}
			for(int k=i; k; k=(k-1)&i){
				if((k >> j) % 2 == 0) continue;
				for(int l=0; l<n; l++){
					dp[i][j] = max(dp[i][j], dp[i^k][l] + adj[l][j] + cyc[k]);
				}
			}
		}
	}
	int ret = dp[(1<<n)-1][n-1];
	for(auto &i : v) ret -= i.x;
	cout << -ret << endl;
}