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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class MaximumWeight {
    int[] values;
    long[][] dp;
    int k;
    ArrayList<Integer>[] adj;
    public static void main(String[] args) throws IOException {
        new MaximumWeight().solve();
    }

    public void solve() throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer tokenizer = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(tokenizer.nextToken());
        k = Integer.parseInt(tokenizer.nextToken()) + 1;
        this.values = new int[n];
        this.dp = new long[n][k + 1];
        this.adj = new ArrayList[n];
        tokenizer = new StringTokenizer(f.readLine());
        for (int i = 0; i < n; i++) {
            values[i] = Integer.parseInt(tokenizer.nextToken());
            adj[i] = new ArrayList<Integer>();
        }
        for (int i = 0; i < n - 1; i++) {
            tokenizer = new StringTokenizer(f.readLine());
            int a = Integer.parseInt(tokenizer.nextToken()) - 1;
            int b = Integer.parseInt(tokenizer.nextToken()) - 1;
            adj[a].add(b);
            adj[b].add(a);
        }
        buildDp(0, -1);
        System.out.println(dp[0][0]);
    }

    public void buildDp(int node, int parent) {
        for (int neigh : adj[node]) {
            if (neigh == parent) {
                continue;
            }
            buildDp(neigh, node);
        }

        for (int i = 1; i <= k; i++) {
            long max = 0;
            for (int to : adj[node]) {
                long current = dp[to][i - 1];
                if (to == parent) {
                    continue;
                }
                for (int neigh : adj[node]) {
                    if (neigh == parent || neigh == to) {
                        continue;
                    }
                    current += dp[neigh][Math.max(k - i - 1, i - 1)];
                }

                max = Math.max(current, max);
            }
            dp[node][i] = max;
        }

        dp[node][0] = values[node] + dp[node][k];
        for (int i = k; i >= 1; i--) {
            dp[node][i - 1] = Math.max(dp[node][i], dp[node][i - 1]);
        }

        /*
        System.out.println(node + 1);
        for (int i = 0; i <= k; i++) {
            System.out.println(dp[node][i] + " " + i);
        }
        System.out.println();

         */
    }
}
