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<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define two(i) (1<<(i-1))
using namespace std;
const int maxn=15+10,inf=1000000000;
int f[(1<<16)+10][maxn],p[(1<<16)+10],id[(1<<16)+10];
int dis[maxn][maxn];
int i,j,k,l,r,s,t,n,m,now,all,ans,sum;
int lowbit(int x){
    return x&-x;
}
int main(){
    scanf("%d%d",&n,&m);
    fo(i,1,n) id[two(i)]=i;
    all=(1<<n)-1;
    fo(i,1,m){
        scanf("%d%d%d",&j,&k,&t);
        sum+=t;
        dis[j][k]=dis[k][j]=t;
    }
    fo(i,1,all){
        now=id[lowbit(i)];
        p[i]=p[i-two(now)];
        fo(j,now+1,n)
            if ((i&two(j))) p[i]+=dis[now][j];
    }
    fo(i,0,n)
        fo(j,0,all)
            f[j][i]=-inf;
    f[two(1)][1]=0;
    ans=-inf;
    fo(s,1,all){
        fo(i,1,n){
            r=s;
            while (r){
                if ((r&two(i))) f[s][i]=max(f[s][i],f[(s-r)|two(i)][i]+p[r]);
                r=(r-1)&s;
            }
            fo(j,1,n)
                if (!(s&two(j))&&dis[i][j]) f[s|two(j)][j]=max(f[s|two(j)][j],f[s][i]+dis[i][j]);
        }
        if ((s&two(n))) ans=max(ans,f[s][n]);
    }
    printf("%d\n",sum-ans);
}