Problem:
Score : 400 points

Problem Statement

There are N towns in the State of Atcoder, connected by M bidirectional roads.

The i-th road connects Town A_i and B_i and has a length of C_i.

Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order).

She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road.

If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?

Constraints

2≤N≤200

1≤M≤N×(N-1)/2

2≤R≤min(8,N) (min(8,N) is the smaller of 8 and N.)

r_i≠r_j (i≠j)

1≤A_i,B_i≤N, A_i≠B_i

(A_i,B_i)≠(A_j,B_j),(A_i,B_i)≠(B_j,A_j) (i≠j)

1≤C_i≤100000

Every town can be reached from every town by road.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M R
r_1 ... r_R
A_1 B_1 C_1
:
A_M B_M C_M

Output

Print the distance traveled by road if Joisino visits the towns in the order that minimizes it.

Sample Input 1

3 3 3
1 2 3
1 2 1
2 3 1
3 1 4

Sample Output 1

2

For example, if she visits the towns in the order of 1, 2, 3, the distance traveled will be 2, which is the minimum possible.

Sample Input 2

3 3 2
1 3
2 3 2
1 3 6
1 2 2

Sample Output 2

4

The shortest distance between Towns 1 and 3 is 4. Thus, whether she visits Town 1 or 3 first, the distance traveled will be 4.

Sample Input 3

4 6 3
2 3 4
1 2 4
2 3 3
4 3 1
1 4 1
4 2 2
3 1 6

Sample Output 3

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma region ヘッダー
#define _CRT_SECURE_NO_WARNINGS

#ifdef _____MY_DEBUG_____
#define VS_ENDPAUSE()	(system("pause"))
#define DEBUG_OUT(...)  (printf(__VA_ARGS__))
#else
#define VS_ENDPAUSE()
#define DEBUG_OUT(...)
#endif

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <functional>

using namespace std;
typedef signed char int8;
typedef signed short int16;
typedef signed int int32;
typedef signed long long int64;
typedef unsigned char uint8;
typedef unsigned short uint16;
typedef unsigned int uint32;
typedef unsigned long long uint64;
#define int int64

#ifndef INT8_MIN
#define INT8_MIN	(-127)
#endif
#ifndef INT8_MAX
#define INT8_MAX	( 128)
#endif
#ifndef UINT8_MAX
#define UINT8_MAX	( 255)
#endif
#ifndef INT16_MIN
#define INT16_MIN	(-32768)
#endif
#ifndef INT16_MAX
#define INT16_MAX	( 32767)
#endif
#ifndef UINT16_MAX
#define UINT16_MAX	( 65535)
#endif
#ifndef INT32_MIN
#define INT32_MIN	(-2147483648)	// -2 * 10^9
#endif
#ifndef INT32_MAX
#define INT32_MAX	( 2147483647)	//  2 * 10^9
#endif
#ifndef UINT32_MAX
#define UINT32_MAX	( 4294967295)	//  4 * 10^9
#endif
#ifndef INT64_MIN
#define INT64_MIN	(-9223372036854775808)	// -9 * 10^18
#endif
#ifndef INT64_MAX
#define INT64_MAX	( 9223372036854775807)	//  9 * 10^18
#endif
#ifndef UINT64_MAX
#define UINT64_MAX	( 18446744073709551615)	// 18 * 10^18
#endif

void func(void);
#pragma endregion

int32 main(void)
{
    func();
    VS_ENDPAUSE();
    return 0;
}

void WarshallFloyd(vector< vector<int32> > &d, int32 n)
{
    for(int32 i=0; i<n; i++){
        for(int32 j=0; j<n; j++){
            for(int32 k=0; k<n; k++){
                d[j][k] = min(d[j][k], d[j][i]+d[i][k]);
            }
        }
    }
}

void func(void)
{
//DEBUG_OUT("----\n");
    int32 N, M, R;
    cin >> N >> M >> R;

    vector<int32> r(R);
    for(int32 i=0; i<R; i++){
        cin >> r[i];
        r[i]--;
    }
    sort(r.begin(), r.end());

    vector< vector<int32> > d(N, vector<int32>(N, INT32_MAX));
    for(int32 i=0; i<N; i++){ d[i][i] = 0; }

    for(int32 i=0; i<M; i++){
        int32 from, to, cost;
        cin >> from >> to >> cost;
        from--; to--;
        d[from][to] = cost;
        d[to][from] = cost;
    }

    WarshallFloyd(d, N);

    int32 ans = INT32_MAX;
    do{
        int32 totalCost = 0;
        for(int32 i=1; i<R; i++){
            int32 from = r[i-1];
            int32 to = r[i];
            totalCost += d[from][to];
        }
        ans = min(ans, totalCost);
    }while(next_permutation(r.begin(), r.end()));

    DEBUG_OUT("----\n");
    cout << ans << endl;

    return;
}