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;

int nth_bit(int64_t num, int n){
    return (num >> n) & 1;
}

int N, M;
int edges[15][15];
int gain[1<<15];

int dp[15][1<<15];
bool visited[15][1<<15];
const int INF = 1e9;

int rec(int start, int bits){
    if(bits == 0){
        return start == N-1 ? 0 : -INF;
    }
    if(visited[start][bits]) return dp[start][bits];

    int mx = -INF;
    for(int i=0; i<N; i++){
        if(nth_bit(bits, i) && edges[start][i]){
            mx = max(mx, rec(i, bits - (1<<i)) + edges[start][i]);
        }
    }

    for(int bits2=bits; bits2>0; bits2=(bits2-1)&bits){
        mx = max(mx, rec(start, bits - bits2) + gain[bits2 + (1<<start)]);
    }
    dp[start][bits] = mx;
    visited[start][bits] = true;
    return mx;
}

int main(){
    cin >> N >> M;
    for(int i=0; i<M; i++){
        int a, b, c;
        cin >> a >> b >> c;
        a--; b--;
        edges[a][b] = c;
        edges[b][a] = c;
        for(int bits=0; bits<(1<<N); bits++){
            if(nth_bit(bits, a) && nth_bit(bits, b)) gain[bits] += c;
        }
    }
    cout << gain[(1<<N)-1] - rec(0, (1<<N)-2) << endl;
    return 0;
}