1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<vector>
 10 #include<set>
 11 
 12 #define N 1005
 13 #define M 100000
 14 #define inf 1000000007
 15 #define mod 1000000007
 16 #define mod2 100000000
 17 #define ll long long
 18 #define maxi(a,b) (a)>(b)? (a) : (b)
 19 #define mini(a,b) (a)<(b)? (a) : (b)
 20 
 21 using namespace std;
 22 
 23 int n;
 24 int m;
 25 int tri[12] ={0,1,3,9,27,81,243,729,2187,6561,19683,59049};
 26 int dig[59050][11];  //dig[state][k_dig]  状态state的第k位是多少
 27 int dp[59050][11];
 28 int d[15][15];
 29 int ans;
 30 int flag;
 31 
 32 void ini1()
 33 {
 34     int o,j,t;
 35     for(o=0;o<59049;o++){
 36         t=o;
 37         for(j=0;j<10;j++){
 38             dig[o][j]=t%3;
 39             t/=3;
 40         }
 41     }
 42 }
 43 
 44 void ini()
 45 {
 46     int a,b;
 47     int c;
 48     int i;
 49     ans=-1;
 50     memset(dp,-1,sizeof(dp));
 51     memset(d,-1,sizeof(d));
 52     while(m--){
 53         scanf("%d%d%d",&a,&b,&c);
 54         if(d[a][b]==-1){
 55             d[a][b]=c;
 56             d[b][a]=c;
 57         }
 58         else{
 59             d[a][b]=min(d[a][b],c);
 60             d[b][a]=min(d[b][a],c);
 61         }
 62     }
 63 
 64     for(i=1;i<=n;i++){
 65         dp[ tri[i] ][i-1]=0;
 66     }
 67 }
 68 
 69 void solve()
 70 {
 71     int o,j,te,k;
 72     for(o=1;o<tri[n+1];o++){
 73         flag=1;
 74         for(j=0;j<n;j++){
 75             if(dig[o][j]==0){
 76                 flag=0;continue;
 77             }
 78             te=o-tri[j+1];
 79             for(k=0;k<n;k++){
 80                 if(dig[te][k]==0) continue;
 81                 if(d[k+1][j+1]==-1) continue;
 82                 if(dp[te][k]==-1) continue;
 83                 if(dp[o][j]==-1){
 84                     dp[o][j]=dp[te][k]+d[k+1][j+1];
 85                 }
 86                 else{
 87                     dp[o][j]=min(dp[o][j],dp[te][k]+d[k+1][j+1]);
 88                 }
 89             }
 90         }
 91 
 92        // printf(" o=%d flag=%d\n",o,flag);
 93         if(flag==0) continue;
 94         for(j=0;j<n;j++){
 95             if(dp[o][j]==-1) continue;
 96             if(ans==-1){
 97                 ans=dp[o][j];
 98             }
 99             else{
100                 ans=min(ans,dp[o][j]);
101             }
102         }
103 
104     }
105 }
106 
107 void out()
108 {
109     //for(int o=0;o<tri[n+1];o++){
110     //    for(int j=0;j<n;j++) printf(" o=%d j=%d dp=%d\n",o,j,dp[o][j]);
111    // }
112     printf("%d\n",ans);
113 }
114 
115 int main()
116 {
117     ini1();
118     //freopen("data.in","r",stdin);
119     //freopen("data.out","w",stdout);
120     //scanf("%d",&T);
121    // for(int cnt=1;cnt<=T;cnt++)
122    // while(T--)
123     while(scanf("%d%d",&n,&m)!=EOF)
124     {
125         ini();
126         solve();
127         out();
128     }
129     return 0;
130 }