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>
//Created by Aminul on 8/30/2019.

import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class F {
    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int maxn = (int) 2e5 + 5;
        ArrayDeque<Node> g[] = genDQ(maxn);
        int n = in.nextInt(), q = in.nextInt();
        for (int i = 0; i < n - 1; i++) {
            int u = in.nextInt(), v = in.nextInt(), w = in.nextInt();
            g[w].addLast(new Node(u, v));
        }
        for (int i = 0; i < q; i++) {
            int w = in.nextInt();
            g[w].addLast(new Node(i));
        }

        DSU dsu = new DSU(n + 1);
        long ans[] = new long[q], paths = 0;
        for (int i = 0; i < maxn; i++) {
            for (Node node : g[i]) {
                if (node.op == 0) {
                    if (dsu.equiv(node.u, node.v)) continue;
                    else {
                        long szu = dsu.getSize(node.u), szv = dsu.getSize(node.v);
                        paths -= (szu * (szu - 1)) / 2;
                        paths -= (szv * (szv - 1)) / 2;
                        dsu.union(node.u, node.v);
                        szu = dsu.getSize(node.u);
                        paths += (szu * (szu - 1)) / 2;
                    }
                } else {
                    ans[node.id] = paths;
                }
            }
        }
        for (long i : ans) pw.print(i + " ");
        pw.println();

        pw.close();
    }

    static <T> ArrayDeque<T>[] genDQ(int n) {
        ArrayDeque<T> list[] = new ArrayDeque[n];
        for (int i = 0; i < n; i++) list[i] = new ArrayDeque<T>();
        return list;
    }

    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }

    static class Node {
        int op, u, v, id;

        Node(int U, int V) {
            op = 0;
            u = U;
            v = V;
        }

        Node(int ID) {
            op = 1;
            id = ID;
        }
    }

    static class DSU {
        int parent[], size[], n;

        DSU(int N) {
            n = N;
            parent = new int[N];
            size = new int[N];
            for (int i = 0; i < N; i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }

        DSU(int[] p) {
            parent = p;
            n = p.length;
        }

        int getSize(int u) {
            return size[find(u)];
        }

        int find(int i) {
            int p = parent[i];
            if (i == p) return i;
            return parent[i] = find(p);
        }

        boolean equiv(int u, int v) {
            return find(u) == find(v);
        }

        boolean union(int u, int v) {
            u = find(u);
            v = find(v);
            if (u == v) return false;
            if (size[u] > size[v]) {
                int t = u;
                u = v;
                v = t;
            }
            parent[u] = v;
            size[v] += size[u];
            return true;
        }

        int count() {
            int cnt = 0;
            for (int i = 0; i < n; i++) {
                if (i == find(i)) cnt++;
            }
            return cnt;
        }
    }
}