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>

using namespace std;

const int INF = 1 << 30;

int N, M;
int g[17][17];
int lookup[1 << 17];
int dp[1 << 17][16];
vector< int > add[1 << 17];

int rec(int mask, int pre)
{
  if(mask == 0) return (0);
  if(~dp[mask][pre]) return (dp[mask][pre]);
  int ret = -INF;
  for(int i = mask; i > 0; i = (i - 1) & mask) {
    for(auto &j : add[i]) {
      if(pre != 15 && g[pre][j] == 0) continue;
      ret = max(ret, rec(mask ^ i, j) + lookup[i] + g[pre][j]);
    }
  }
  return (dp[mask][pre] = ret);
}

signed main()
{
  memset(dp, -1, sizeof(dp));

  cin >> N >> M;
  int all = 0;

  for(int i = 0; i < M; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    --a, --b;
    g[a][b] = c;
    g[b][a] = c;
    all += c;
  }

  for(int i = 0; i < 1 << N; i++) {
    if(i & 1 && (i >> (N - 1)) & 1) continue;
    for(int j = 0; j < N; j++) {
      if((i >> j) & 1) {
        if(i & 1) {
          if(j == 0) add[i].push_back(j);
        } else if((i >> (N - 1)) & 1) {
          if(j == N - 1) add[i].push_back(j);
        } else {
          add[i].push_back(j);
        }
        for(int k = j + 1; k < N; k++) {
          if((i >> k) & 1) lookup[i] += g[j][k];
        }
      }
    }
  }
  cout << all - rec((1 << N) - 1, 15) << endl;
}