Chouti was tired of the tedious homework, so he opened up an old programming problem he created years ago.

You are given a connected undirected graph with n vertices and m weighted edges. There are k special vertices: x_1, x_2, …, x_k.

Let's define the cost of the path as the maximum weight of the edges in it. And the distance between two vertexes as the minimum cost of the paths connecting them.

For each special vertex, find another special vertex which is farthest from it (in terms of the previous paragraph, i.e. the corresponding distance is maximum possible) and output the distance between them.

The original constraints are really small so he thought the problem was boring. Now, he raises the constraints and hopes you can solve it for him.

Input

The first line contains three integers n, m and k (2 ≤ k ≤ n ≤ 10^5, n-1 ≤ m ≤ 10^5) — the number of vertices, the number of edges and the number of special vertices.

The second line contains k distinct integers x_1, x_2, …, x_k (1 ≤ x_i ≤ n).

Each of the following m lines contains three integers u, v and w (1 ≤ u,v ≤ n, 1 ≤ w ≤ 10^9), denoting there is an edge between u and v of weight w. The given graph is undirected, so an edge (u, v) can be used in the both directions.

The graph may have multiple edges and self-loops.

It is guaranteed, that the graph is connected.

Output

The first and only line should contain k integers. The i-th integer is the distance between x_i and the farthest special vertex from it.

Examples

Input

2 3 2
2 1
1 2 3
1 2 2
2 2 1


Output

2 2 


Input

4 5 3
1 2 3
1 2 5
4 2 1
2 3 2
1 4 4
1 3 3


Output

3 3 3 

Note

In the first example, the distance between vertex 1 and 2 equals to 2 because one can walk through the edge of weight 2 connecting them. So the distance to the farthest node for both 1 and 2 equals to 2.

In the second example, one can find that distance between 1 and 2, distance between 1 and 3 are both 3 and the distance between 2 and 3 is 2.

The graph may have multiple edges between and self-loops, as in the first example.
4 4 3
1 2 3
1 2 1
2 3 2
1 3 3
1 4 4
 4 3 2
1 4
1 2 1
2 3 5
3 4 1
 3 2 2
1 2
1 2 1
2 3 47
 3 2 2
1 2
1 2 1
2 3 1000
 3 2 2
1 2
1 2 10
2 3 100
 3 3 2
2 3
1 2 100
1 3 100
2 3 1
 3 2 2
1 2
1 2 2
2 3 3
 6 5 3
1 2 4
1 3 3
3 2 2
2 4 1
3 5 4
5 6 10
 3 2 2
2 3
1 2 10
2 3 1
 3 2 2
1 2
1 2 1
2 3 4
 6 5 2
1 6
1 2 1
2 3 2
3 4 3
4 5 2
5 6 1
 5 4 2
4 5
1 2 100
2 3 100
3 4 10
3 5 20
 4 3 3
1 2 3
1 2 5
1 3 4
1 4 5
 4 4 3
1 2 3
1 2 1
1 3 2
2 3 3
3 4 5
 2 2 2
1 2
1 2 3
1 2 5
 3 2 2
1 2
1 2 1
2 3 100
 3 3 2
1 2
1 2 1
2 3 4
1 3 5
 3 2 2
2 3
1 2 3
2 3 1
 5 4 2
4 5
1 2 10
2 3 10
3 4 1
4 5 1
 4 3 2
1 4
1 2 4
2 3 6
3 4 4
 4 3 2
1 2
1 2 1
2 3 23
3 4 1231
 3 2 2
1 2
1 2 3
2 3 5
 5 4 2
1 5
1 2 1
1 3 2
2 4 5
3 5 3
 3 4 2
2 1
1 2 3
1 2 2
2 2 1
1 3 99
 4 3 2
1 2
1 2 1
2 3 123
3 4 12321
 4 3 3
1 2 3
1 2 6
1 3 7
1 4 10
 4 5 2
2 3
1 2 5
4 2 1
2 3 2
1 4 4
1 3 3
 3 2 2
1 2
1 2 1
2 3 5
 3 2 2
1 2
1 2 1
2 3 2
 3 2 2
1 2
3 2 10
2 1 1
 4 3 2
3 4
1 2 2
1 3 4
3 4 1
 3 2 2
1 3
1 2 1
2 3 1
 5 5 2
1 2
1 2 1
2 3 2
3 4 2
4 5 2
5 1 2
 5 4 3
1 2 4
1 2 10
2 3 100
2 4 20
5 3 1000
 4 3 2
3 4
1 2 9
2 3 6
3 4 1
 3 2 2
1 2
1 2 10
3 2 20
 5 4 2
4 5
1 2 100
2 3 100
3 4 100
4 5 1
 3 2 2
3 2
1 2 233
2 3 3
 4 3 2
1 4
1 2 1
2 3 3
3 4 1
 2 1 2
1 2
1 2 1000000000
 5 7 4
1 2 3 4
1 2 3
5 1 4
3 1 1
4 2 5
2 5 6
2 3 3
3 4 6
 4 3 2
1 2
1 2 1
2 3 2
3 4 1
 3 2 2
2 3
1 2 1000
2 3 1
 4 3 2
3 4
1 2 10000
2 3 10000
3 4 1
 7 6 2
6 7
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1
6 7 1
 3 2 2
2 3
1 2 100
2 3 1
 4 3 2
1 2
1 2 1
2 3 12
3 4 123123
 3 2 2
2 1
1 2 1
2 3 100
 3 2 2
2 3
1 2 2
2 3 1
 3 2 2
2 3
1 2 7
2 3 1
 3 2 2
1 2
1 2 1
2 3 3
 3 2 2
1 2
1 2 10
2 3 15
 4 3 2
1 4
1 2 3
2 3 4
3 4 3
 6 5 4
1 2 3 4
1 2 1
2 3 1
3 4 1
4 5 1
5 6 10
 4 3 2
2 3
1 2 1000
2 3 1
3 4 1000
 4 3 2
1 2
1 2 1
2 3 1000
3 4 1000
 3 3 2
1 2
1 2 1
1 3 1000
2 3 1000
 3 2 2
1 2
1 2 3
1 3 5
 4 4 2
3 4
1 2 1000000000
2 3 1000000000
3 1 1000000000
3 4 1
 4 3 2
3 4
1 2 5
2 3 3
2 4 4
 3 2 2
1 2
1 2 3
2 3 100
 3 2 2
2 3
1 2 5
2 3 1
 5 4 2
4 5
1 2 10
2 3 10
3 4 1
3 5 1
 4 3 2
2 3
1 2 100
2 3 1
3 4 100
 4 3 3
1 2 3
1 2 1
3 1 2
4 3 3
 4 3 2
3 4
1 2 100
2 3 2
2 4 2

2 2 2  5 5  1 1  1 1  10 10  1 1  2 2  3 3 3  1 1  1 1  3 3  20 20  5 5 5  2 2 2  3 3  1 1  1 1  1 1  1 1  6 6  1 1  3 3  3 3  2 2  1 1  7 7 7  2 2  1 1  1 1  1 1  1 1  1 1  1 1  20 20 20  1 1  10 10  1 1  3 3  3 3  1000000000 1000000000  5 5 5 5  1 1  1 1  1 1  1 1  1 1  1 1  1 1  1 1  1 1  1 1  10 10  4 4  1 1 1 1  1 1  1 1  1 1  3 3  1 1  4 4  3 3  1 1  1 1  1 1  2 2 2  2 2 
import java.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class cf1018d_2 {
    static BufferedReader __in;
    static PrintWriter __out;
    static StringTokenizer input;

    static int k, par[], sz[], cnt[];
    static boolean end = false;

    static int find(int i) {
        return i == par[i] ? i : (par[i] = find(par[i]));
    }

    static void union(int i, int j, int w) {
        int a = find(i), b = find(j);
        if(a != b) {
            if(sz[b] > sz[a]) {
                int swap = a;
                a = b;
                b = swap;
            }
            par[b] = a;
            sz[a] += sz[b];
            cnt[a] += cnt[b];
            if(cnt[a] == k) {
                end = true;
                for(int l = 0; l < k; ++l) {
                    pr(w);
                    pr(' ');
                }
                prln();
            }
        }
    }

    public static void main(String[] args) throws IOException {
        __in = new BufferedReader(new InputStreamReader(System.in));
        __out = new PrintWriter(new OutputStreamWriter(System.out));

        int n = rni(), m = ni();
        k = ni();
        par = new int[n];
        sz = new int[n];
        cnt = new int[n];
        for(int i = 0; i < n; ++i) {
            par[i] = i;
            sz[i] = 1;
        }
        r();
        for(int i = 0; i < k; ++i) {
            cnt[ni() - 1] = 1;
        }
        int[][] e = new int[m][3];
        for(int i = 0; i < m; ++i) {
            e[i][0] = rni() - 1;
            e[i][1] = ni() - 1;
            e[i][2] = ni();
        }
        sort(e, (a, b) -> a[2] - b[2]);
        int ind = 0;
        while(!end) {
            union(e[ind][0], e[ind][1], e[ind++][2]);
        }

        close();
    }

    // references
    // IBIG = 1e9 + 7
    // IRAND ~= 3e8
    // IMAX ~= 2e10
    // LMAX ~= 9e18
    // constants
    static final int IBIG = 1000000007;
    static final int IRAND = 327859546;
    static final int IMAX = 2147483647;
    static final int IMIN = -2147483648;
    static final long LMAX = 9223372036854775807L;
    static final long LMIN = -9223372036854775808L;
    // util
    static int minof(int a, int b, int c) {return min(a, min(b, c));}
    static int minof(int... x) {return x.length == 1 ? x[0] : x.length == 2 ? min(x[0], x[1]) : min(x[0], minstarting(1, x));}
    static int minstarting(int offset, int... x) {assert x.length > 2; return offset == x.length - 2 ? min(x[offset], x[offset + 1]) : min(x[offset], minstarting(offset + 1, x));}
    static long minof(long a, long b, long c) {return min(a, min(b, c));}
    static long minof(long... x) {return x.length == 1 ? x[0] : x.length == 2 ? min(x[0], x[1]) : min(x[0], minstarting(1, x));}
    static long minstarting(int offset, long... x) {assert x.length > 2; return offset == x.length - 2 ? min(x[offset], x[offset + 1]) : min(x[offset], minstarting(offset + 1, x));}
    static int maxof(int a, int b, int c) {return max(a, max(b, c));}
    static int maxof(int... x) {return x.length == 1 ? x[0] : x.length == 2 ? max(x[0], x[1]) : max(x[0], maxstarting(1, x));}
    static int maxstarting(int offset, int... x) {assert x.length > 2; return offset == x.length - 2 ? max(x[offset], x[offset + 1]) : max(x[offset], maxstarting(offset + 1, x));}
    static long maxof(long a, long b, long c) {return max(a, max(b, c));}
    static long maxof(long... x) {return x.length == 1 ? x[0] : x.length == 2 ? max(x[0], x[1]) : max(x[0], maxstarting(1, x));}
    static long maxstarting(int offset, long... x) {assert x.length > 2; return offset == x.length - 2 ? max(x[offset], x[offset + 1]) : max(x[offset], maxstarting(offset + 1, x));}
    static int powi(int a, int b) {if(a == 0) return 0; int ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static long powl(long a, int b) {if(a == 0) return 0; long ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static int floori(double d) {return (int)d;}
    static int ceili(double d) {return (int)ceil(d);}
    static long floorl(double d) {return (long)d;}
    static long ceill(double d) {return (long)ceil(d);}
    // input
    static void r() throws IOException {input = new StringTokenizer(__in.readLine());}
    static int ri() throws IOException {return Integer.parseInt(__in.readLine());}
    static long rl() throws IOException {return Long.parseLong(__in.readLine());}
    static int[] ria(int n) throws IOException {int[] a = new int[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Integer.parseInt(input.nextToken()); return a;}
    static long[] rla(int n) throws IOException {long[] a = new long[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Long.parseLong(input.nextToken()); return a;}
    static char[] rcha() throws IOException {return __in.readLine().toCharArray();}
    static String rline() throws IOException {return __in.readLine();}
    static int rni() throws IOException {input = new StringTokenizer(__in.readLine()); return Integer.parseInt(input.nextToken());}
    static int ni() {return Integer.parseInt(input.nextToken());}
    static long rnl() throws IOException {input = new StringTokenizer(__in.readLine()); return Long.parseLong(input.nextToken());}
    static long nl() {return Long.parseLong(input.nextToken());}
    // output
    static void pr(int i) {__out.print(i);}
    static void prln(int i) {__out.println(i);}
    static void pr(long l) {__out.print(l);}
    static void prln(long l) {__out.println(l);}
    static void pr(double d) {__out.print(d);}
    static void prln(double d) {__out.println(d);}
    static void pr(char c) {__out.print(c);}
    static void prln(char c) {__out.println(c);}
    static void pr(char[] s) {__out.print(new String(s));}
    static void prln(char[] s) {__out.println(new String(s));}
    static void pr(String s) {__out.print(s);}
    static void prln(String s) {__out.println(s);}
    static void pr(Object o) {__out.print(o);}
    static void prln(Object o) {__out.println(o);}
    static void prln() {__out.println();}
    static void pryes() {__out.println("yes");}
    static void pry() {__out.println("Yes");}
    static void prY() {__out.println("YES");}
    static void prno() {__out.println("no");}
    static void prn() {__out.println("No");}
    static void prN() {__out.println("NO");}
    static void pryesno(boolean b) {__out.println(b ? "yes" : "no");};
    static void pryn(boolean b) {__out.println(b ? "Yes" : "No");}
    static void prYN(boolean b) {__out.println(b ? "YES" : "NO");}
    static void prln(int... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
    static void prln(long... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
    static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for(int i = 0; i < n; __out.print(iter.next()), __out.print(' '), ++i); __out.println(iter.next());}
    static void flush() {__out.flush();}
    static void close() {__out.close();}
}