p03664 AtCoder Regular Contest 078 - Mole and Abandoned Mine
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
```python
n, m = map(int, input().split())

g = [[0 for j in range(n)] for i in range(n)]

for i in range(m):
    u, v, w = map(int, input().split())
    g[u - 1][v - 1] = g[v - 1][u - 1] = w

e = [sum(g[i][j] for i in range(n) if S >> i & 1 for j in range(i + 1, n) if S >> j & 1) for S in range(1 << n)]

dp = [[-10 ** 9 for j in range(n)] for i in range(1 <<  n)]

for i in range(1 << n):
    for j in range(n):
        if i >> j & 1:
            if not j:
                dp[i][j] = e[i]
            else:
                for k in range(n):
                    if  j != k and (i >> k & 1) and g[k][j]:
                        dp[i][j] = max(dp[i][j], dp[i ^ (1 << j)][k] + g[k][j]) 
                s = i ^ (1 << j)
                k = s
                while k:
                    dp[i][j] = max(dp[i][j], dp[i ^ k][j] + e[k | 1 << j])
                    k = (k - 1) & s 

print(e[(1 << n) - 1] - dp[(1 << n) - 1][n - 1])
```