Problem Description

After trying hard for many years, Victor has finally received a pilot license. To have a celebration, he intends to buy himself an airplane and fly around the world. There are n countries on the earth, which are numbered from 1 to n. They are connected by m undirected flights, detailedly the i-th flight connects the ui-th and the vi-th country, and it will cost Victor's airplane wi L fuel if Victor flies through it. And it is possible for him to fly to every country from the first country.

Victor now is at the country whose number is 1, he wants to know the minimal amount of fuel for him to visit every country at least once and finally return to the first country.

 

Input

The first line of the input contains an integer T, denoting the number of test cases.
In every test case, there are two integers n and m in the first line, denoting the number of the countries and the number of the flights.

Then there are m lines, each line contains three integers ui, vi and wi, describing a flight.

1≤T≤20.

1≤n≤16.

1≤m≤100000.

1≤wi≤100.

1≤ui,vi≤n.

 

Output

Your program should print T lines : the i-th of these should contain a single integer, denoting the minimal amount of fuel for Victor to finish the travel.

 

Sample Input

 

Sample Output

 

#include <bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define LC(x) (x<<1)

#define RC(x) ((x<<1)+1)

#define MID(x,y) ((x+y)>>1)

#define CLR(arr,val) memset(arr,val,sizeof(arr))

#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);

typedef pair<int, int> pii;

typedef long long LL;

const double PI = acos(-1.0);

const int N = 17;

int E[N][N], dp[1 << N][N];

void init()

{

    CLR(E, INF);

    CLR(dp, INF);

    for (int i = 0; i < N; ++i)

        E[i][i] = 0;

}

int main(void)

{

    int tcase, n, m, i;

    scanf("%d", &tcase);

    while (tcase--)

    {

        init();

        scanf("%d%d", &n, &m);

        for (i = 0; i < m; ++i)

        {

            int a, b, w;

            scanf("%d%d%d", &a, &b, &w);

            --a;

            --b;

            E[a][b] = E[b][a] = min(E[a][b], w);

        }

        for (int k = 0; k < n; ++k)

            for (int i = 0; i < n; ++i)

                for (int j = 0; j < n; ++j)

                    E[i][j] = min(E[i][j], E[i][k] + E[k][j]);

        dp[0][0] = 0;

        int st_cnt = 1 << n;

        for (int s = 0; s < st_cnt; ++s) //已经走过的点的状态

        {

            for (int u = 0; u < n; ++u)

            {

                if (dp[s][u] != INF) //选取其中实际可行的方案

                {

                    for (int v = 0; v < n; ++v) //枚举下一步走的方案

                    {

                        if ((s & (1 << v)))//v如果已经走过了就没有必要再走

                            continue;

                        int news = s | (1 << v);

                        dp[news][v] = min(dp[news][v], dp[s][u] + E[u][v]);

                    }

                }

            }

        }

        printf("%d\n", dp[(1 << n) - 1][0]);

    }

    return 0;

}