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>
#define maxn 18

using namespace std;

int n, m;
int a[maxn][maxn];
int x, y, z;
int g[1 << maxn];
int f[1 << maxn][maxn];
int sum;

vector<int> v;

inline void Max(int &x, int y){
	x = max(x, y);
}

int main(){
	scanf("%d%d", &n, &m);
	while(m--){
		scanf("%d%d%d", &x, &y, &z);
		--x, --y;
		a[x][y] = a[y][x] = z;
		sum += z;
	}
	for(int i = 0;i < (1 << n);i++){
		v.clear();
		for(int j = 0;j < n;j++){
			if(i & (1 << j)) v.push_back(j);
		}
		for(int j = 1;j < v.size();j++){
			for(int k = 0;k < j;k++){
				g[i] += a[v[j]][v[k]];//debug:没写v 我弱智 
			}
		}
	}
	memset(f, -0x3f, sizeof(f));
	f[1][0] = 0;
	for(int i = 0;i < (1 << n) - 1;i++){
		for(int j = 0;j < n;j++){
			if(!(i & (1 << j))) continue;
			for(int k = 0;k < n;k++){
				if(!(i & (1 << k)) && a[j][k]){
					Max(f[i | (1 << k)][k], f[i][j] + a[j][k]);
				}
			}
			int x = i ^ ((1 << n) - 1);
			for(int k = x;k;k = (k - 1) & x){
				Max(f[i | k][j], f[i][j] + g[(1 << j) | k]);
			}
		}	
	}
	int ans = 0;
	for(int i = 0;i < (1 << n);i++) ans = max(ans, f[i][n - 1]);
	printf("%d", sum - ans);
}