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;

#define fo(i,s,t) for(int i = s; i <= t; ++ i)
const int maxn = 15;

int n, m, f[1<<maxn][maxn], cost[1<<maxn], dis[maxn][maxn];

int calc(int s,int t){return cost[s|t]-cost[s]-cost[t];}
void upd(int &a,int b){a = (a < b ? a : b);}
int main()
{
	scanf("%d%d",&n,&m);
	fo(i,1,m)
	{
		int x, y, z; scanf("%d%d%d",&x,&y,&z);
		-- x; -- y; dis[x][y] = dis[y][x] = z;
	}
	fo(i,0,(1<<n)-1)
		fo(j,0,n-1) if(i&(1<<j))
			fo(k,j+1,n-1) if((i&(1<<k))&&dis[j][k])
				cost[i] += dis[j][k];
	memset(f,0x3f,sizeof(f));
	f[1][0] = 0;
	fo(s,0,(1<<n)-1) fo(i,0,n-1) if(f[s][i]!=0x3f3f3f3f&&(s&(1<<i)))
	{
		fo(j,0,n-1) if((!(s&(1<<j)))&&dis[i][j])
			upd(f[s|(1<<j)][j], f[s][i] + calc(s-(1<<i),(1<<j)));
		int t = (1<<n)-1-s;
		for(int y = t; y >= 1; y = ((y-1)&t))
			upd(f[s|y][i], f[s][i] + calc(s-(1<<i),y));
	}
	printf("%d\n",f[(1<<n)-1][n-1]);
	return 0;
}