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 <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#define rep(i,x,y) for (int i=x; i<=y; ++i)
#define repd(i,x,y) for (int i=x; i>=y; --i)

int get()
{
    char c;
    while (!isdigit(c=getchar()) && c!='-');
    int s=c=='-'? -1:1,k=~s? c-'0':0;
    for (; isdigit(c=getchar()); k=k*10+c-'0');
    return s*k;
}

using namespace std;
typedef long long ll;
const int N=15,S=1<<N|1;
int n,m,s,tot,w[N][N],f[S][N],g[S];

int main()
{
    n=get(),m=get(),s=(1<<n)-1;
    rep(i,1,m)
    {
        int u=get()-1,v=get()-1;
        tot+=w[u][v]=w[v][u]=get();
    }
    rep(i,1,s)
    {
        int x=0;
        for (; !((i>>x)&1); ++x);
        g[i]=g[i^(1<<x)];
        rep(j,x+1,n-1)
            if ((i>>j)&1)
                g[i]+=w[x][j];
    }
    memset(f,128,sizeof(f)),f[0][0]=0;
    rep(k,1,s)
    {
        rep(i,0,n-1)
            if ((k>>i)&1)
                for (int _=k^(1<<i),j=_; ~j; ~--j? j&=_:0)
                    f[k][i]=max(f[k][i],f[j][i]+g[k^j]);
        rep(i,0,n-1)
            if (!((k>>i)&1))
                rep(j,0,n-1)
                    if ((k>>j)&1 && w[j][i])
                        f[k][i]=max(f[k][i],f[k][j]+w[j][i]);
    }
    printf("%d",tot-f[s][n-1]);
    return 0;
}