1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 #define ms(a,b) memset((a),(b),sizeof((a)))
13 using namespace std;
14 typedef long long LL;
15 const double EPS = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 2e18;
18 const int MAXN = 10+10;
19 
20 int n, m;
21 int g[MAXN][MAXN], c[MAXN], dp[200000][10];
22 
23 int main()
24 {
25     c[0] = 1;
26     for(int i = 1; i<=10; i++)  //计算三进制
27         c[i] = c[i-1]*3;
28 
29     while(scanf("%d%d", &n, &m)!=EOF)
30     {
31         memset(g, -1, sizeof(g));
32         for(int i = 1; i<=m; i++)
33         {
34             int u, v, w;
35             scanf("%d%d%d", &u, &v, &w);
36             u--; v--;
37             if(g[u][v]!=-1) w = min(w, g[u][v]);
38             g[u][v] = g[v][u] = w;
39         }
40 
41         memset(dp, -1, sizeof(dp));
42         for(int i = 0; i<n; i++)
43             dp[c[i]][i] = 0;
44 
45         int ans = INF;
46         for(int s = 1; s<c[n]; s++)
47         for(int j = 0; j<n; j++)
48         {
49             if(dp[s][j]==-1) continue;  //此状态不存在， 跳过
50 
51             int cnt = 0;
52             for(int k = 0; k<n; k++)
53             {
54                 if(j!=k && g[j][k]!=-1 && (s/c[k])%3<2 )
55                 {
56                     int tmp = s + c[k];
57                     if(dp[tmp][k]==-1 || dp[tmp][k]>dp[s][j]+g[j][k])
58                         dp[tmp][k] = dp[s][j] + g[j][k];
59                 }
60 
61                 if((s/c[k])%3) cnt++;   //统计状态s下走了多少个点
62             }
63 
64             if(cnt==n) ans = min(ans, dp[s][j]);    //如果状态s下走完所有点， 则更新答案
65         }
66 
67         printf("%d\n", ans==INF?-1:ans);
68     }
69 }