#include <fstream>
#include <queue>
#include <vector>
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

using namespace std;

class InParser {
private:
	FILE *fin;
	char *buff;
	int sp;

	char read_ch() {
		++sp;
		if (sp == 4096) {
			sp = 0;
			fread(buff, 1, 4096, fin);
		}
		return buff[sp];
	}

public:
	InParser(const char* nume) {
		fin = fopen(nume, "r");
		buff = new char[4096]();
		sp = 4095;
	}

	InParser& operator >> (int &n) {
		char c;
		while (!isdigit(c = read_ch()) && c != '-');
		int sgn = 1;
		if (c == '-') {
			n = 0;
			sgn = -1;
		} else {
			n = c - '0';
		}
		while (isdigit(c = read_ch())) {
			n = 10 * n + c - '0';
		}
		n *= sgn;
		return *this;
	}

	InParser& operator >> (long long &n) {
		char c;
		n = 0;
		while (!isdigit(c = read_ch()) && c != '-');
		long long sgn = 1;
		if (c == '-') {
			n = 0;
			sgn = -1;
		} else {
			n = c - '0';
		}
		while (isdigit(c = read_ch())) {
			n = 10 * n + c - '0';
		}
		n *= sgn;
		return *this;
	}
} f("ubuntzei.in");

class OutParser {
private:
    FILE *fout;
    char *buff;
    int sp;

    void write_ch(char ch) {
        if (sp == 50000) {
            fwrite(buff, 1, 50000, fout);
            sp = 0;
            buff[sp++] = ch;
        } else {
            buff[sp++] = ch;
        }
    }


public:
    OutParser(const char* name) {
        fout = fopen(name, "w");
        buff = new char[50000]();
        sp = 0;
    }
    ~OutParser() {
        fwrite(buff, 1, sp, fout);
        fclose(fout);
    }

    OutParser& operator << (int vu32) {
        if (vu32 <= 9) {
            write_ch(vu32 + '0');
        } else {
            (*this) << (vu32 / 10);
            write_ch(vu32 % 10 + '0');
        }
        return *this;
    }

    OutParser& operator << (long long vu64) {
        if (vu64 <= 9) {
            write_ch(vu64 + '0');
        } else {
            (*this) << (vu64 / 10);
            write_ch(vu64 % 10 + '0');
        }
        return *this;
    }

    OutParser& operator << (char ch) {
        write_ch(ch);
        return *this;
    }
    OutParser& operator << (const char *ch) {
        while (*ch) {
            write_ch(*ch);
            ++ch;
        }
        return *this;
    }
} g("ubuntzei.out");

int n, m, k;

priority_queue < pair <int, int>, vector <pair <int, int> >, greater <pair <int, int> > > q;

vector <int> loc, cost;

vector <vector <pair <int, int> > > a;

vector <bool> viz;

void dijkstra(int popas)
{
    for (int i = 1; i <= n; ++i) {
        viz[i] = 0;
        cost[i] = 1 << 30;
    }

    cost[popas] = 0;
    q.push({0, popas});

    while (!q.empty()) {
        pair <int, int> nod = q.top();

        q.pop();

        if (viz[nod.second]) {
            continue;
        }

        viz[nod.second] = 1;

        for (auto vecin : a[nod.second]) {
            if (vecin.second + nod.first < cost[vecin.first]) {
                cost[vecin.first] = vecin.second + nod.first;
                q.push({cost[vecin.first], vecin.first});
            }
        }
    }
}

int main()
{
    f >> n >> m >> k;

    a.resize(n + 1), loc.resize(k + 2), viz.resize(n + 1, 0), cost.resize(n + 1, (1 << 30));

    for (int i = 1; i <= k; ++i) {
        int x;
        f >> x;
        loc[i] = x;
    }

    for (int i = 1; i <= m; ++i) {
        int x, y, z;

        f >> x >> y >> z;

        a[x].push_back({y, z});
        a[y].push_back({x, z});
    }

    loc[0] = 1;
    ++k;
    loc[k] = n;

    vector <vector <int> > matCost(k + 1, vector <int> (k + 1, (1 << 30)));

    for (int i = 0; i <= k; ++i) {
        dijkstra(loc[i]);

        for (int j = 0; j <= k; ++j) {
            matCost[i][j] = matCost[j][i] = cost[loc[j]];
        }
    }

    vector <vector <int> > dp((1 << (k + 1)), vector <int> (k + 1, (1 << 30)));

    dp[0][0] = 0;

    for (int mask = 0; mask < (1 << (k + 1)); ++mask) {
        for (int i = 0; i <= k; ++i) {
            if (mask & (1 << i)) {
                for (int j = 0; j <= k; ++j) {
                    dp[mask][i] = min(dp[mask][i], dp[mask ^ (1 << i)][j] + matCost[i][j]);
                }
            }
        }
    }

    g << dp[(1 << (k + 1)) - 1][k];

    return 0;
}