You are given a weighted tree consisting of n vertices. Recall that a tree is a connected graph without cycles. Vertices u_i and v_i are connected by an edge with weight w_i.

You are given m queries. The i-th query is given as an integer q_i. In this query you need to calculate the number of pairs of vertices (u, v) (u < v) such that the maximum weight of an edge on a simple path between u and v doesn't exceed q_i.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of vertices in the tree and the number of queries.

Each of the next n - 1 lines describes an edge of the tree. Edge i is denoted by three integers u_i, v_i and w_i — the labels of vertices it connects (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i) and the weight of the edge (1 ≤ w_i ≤ 2 ⋅ 10^5). It is guaranteed that the given edges form a tree.

The last line of the input contains m integers q_1, q_2, ..., q_m (1 ≤ q_i ≤ 2 ⋅ 10^5), where q_i is the maximum weight of an edge in the i-th query.

Output

Print m integers — the answers to the queries. The i-th value should be equal to the number of pairs of vertices (u, v) (u < v) such that the maximum weight of an edge on a simple path between u and v doesn't exceed q_i.

Queries are numbered from 1 to m in the order of the input.

Examples

Input


7 5
1 2 1
3 2 3
2 4 1
4 5 2
5 7 4
3 6 2
5 2 3 4 1


Output


21 7 15 21 3 


Input


1 2
1 2


Output


0 0 


Input


3 3
1 2 1
2 3 2
1 3 2


Output


1 3 3 

Note

The picture shows the tree from the first example: <image>
import java.util.*;

public class cf1249F {

    static int n, k;
    static int dp[][], a[];
    static ArrayList<Integer> edge[];

    static void init () {
        dp = new int[n+1][n+1];
        a = new int[n+1];

        edge = new ArrayList[n+1];
        for (int i=0; i<=n; i++) {
            edge[i] = new ArrayList<Integer>();
        }
    }

    static void dfs (int nd, int par, int dep) {

        dp[nd][dep] = a[nd];
        for (int v : edge[nd]) {
            if (v != par) {
                dfs(v, nd, dep+1);
                int d2 = dep+k+1;
                if (d2 <= n) {
                    dp[nd][dep] += dp[v][d2];
                }
            }
        }
        
        // System.out.println(nd + ", " + dep + ", " + dp[nd][dep]);
        for (int d=dep+1; d<=n; d++) {
            dp[nd][d] = 0;
            for (int c : edge[nd]) {
                int cur = dp[c][d];

                for (int v : edge[nd]) {
                    if (v != c) {
                        int d2 = Math.max(d, 1+k-(d-dep)+dep);
                        if (d2 <= n) {
                            cur += dp[v][d2];
                        }
                    }
                }

                dp[nd][d] = Math.max(dp[nd][d], cur);
            }

            // System.out.println(nd + ", " + d + ", " + dp[nd][d]);
        }

        for (int i=n-1; i>=dep; i--) {
            dp[nd][i] = Math.max(dp[nd][i], dp[nd][i+1]);
        }
    }

    public static void main (String[] args) {

        Scanner in = new Scanner(System.in);

        n = in.nextInt();
        k = in.nextInt();

        init();

        for (int i=1; i<=n; i++) {
            a[i] = in.nextInt();
        }

        for (int i=0; i<n-1; i++) {
            int u, v;
            u = in.nextInt();
            v = in.nextInt();

            edge[u].add(v);
            edge[v].add(u);
        }

        dfs(1, -1, 0);

        System.out.println(dp[1][0]);

        in.close();
    }
}