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 JAVA):

Solution:
import java.util.*;

public class Main {

	public static void main(String[] args) {
		
		int[][] dp=new int [1<<20][20];
		Scanner sc=new Scanner(System.in);
		int n=(int) sc.nextLong();
		int m=sc.nextInt();
		int sum=0;
		int distance[][]=new int[n][n];
		for (int i = 0; i < m; i++) {
			int u=sc.nextInt();
			int v=sc.nextInt();
			int c=sc.nextInt();
			u--;v--;
			distance[u][v]=c;
			distance[v][u]=c;
			sum+=c;
		}
		int [] s=new int[1<<n];
		for (int i = 1; i <=1<<n; i++) {
			for (int j = 0; j <n; j++) {
				for (int k = j+1; k <n; k++) {
					if((i>>j)%2==1&&(i>>k)%2==1)
					{
						s[i]+=distance[j][k];
					}
				}
			}
			
		}
		//System.out.println(Arrays.toString(s));
		for (int i = 0; i <= 1<<n; i++) {
			Arrays.fill(dp[i], -1);
		}
		dp[1][0]=0;
		for (int i = 0; i <(1<<n); i++) {
			for (int j = 0; j <n; j++) {
				if(dp[i][j]==-1)
					continue;
				for (int k = 0; k < n; k++) {					
					if((i>>k)%2==1||distance[j][k]==0)
						continue;
					dp[i|(1<<k)][k]=Math.max(dp[i|(1<<k)][k], dp[i][j]+distance[j][k]);
						
				}
				int x=(1<<n)-1-i;
				for (int y=x;y>=1;y=(y-1)&x)
					dp[i|y][j]=Math.max(dp[i|y][j],dp[i][j]+s[y|(1<<j)]);
			}
		}
		System.out.println(sum-dp[(1<<n)-1][n-1]);
	
	}

}