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 ll;

template <typename _tp> inline void read(_tp&x){
	char ch=getchar(),ob=0;x=0;
	while(ch!='-'&&!isdigit(ch))ch=getchar();if(ch=='-')ob=1,ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();if(ob)x=-x;
}

#define bin(x) (1<<(x))
#define b(x) (1<<(x)>>1)

inline void cmax(int&x, const int y) {x < y && (x = y);}

const int N = 17, M = 40100;
int f[N][M], g[M];
int mp[N][N];
int n, m;

int main() {
	read(n), read(m);
	int Ans = 0;
	for(int i=1,x,y,z;i<=m;++i) {
		read(x), read(y), read(z);
		Ans += z;
		mp[x][y] = mp[y][x] = z;
	}
	for(int s=0;s<bin(n);++s) {
		int&v = g[s];
		for(int i=1;i<=n;++i) if(s & b(i)) 
		for(int j=i+1;j<=n;++j) if(s & b(j)) v += mp[i][j];
	}
	for(int i=1;i<=n;++i)
	for(int j=0;j<bin(n);++j)
		f[i][j] = -1;
	
	f[1][1] = 0;
	for(int S=1;S<bin(n);++S)
	for(int i=1;i<=n;++i) if((S & b(i)) and ~f[i][S]) {
		for(int j=1;j<=n;++j) if((~S & b(j)) and mp[i][j]) cmax(f[j][S|b(j)], f[i][S] + mp[i][j]);
		for(int iS=(bin(n)-1)^S, s = iS; s; s = (s-1) & iS)
			cmax(f[i][S|s], f[i][S] + g[s|b(i)]);
	}
	printf("%d\n", Ans - f[n][bin(n)-1]);
	return 0;
}