Problem: 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>

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class G {

	static class UnionFind {

		private int[] parent;
		private int[] rank;

		public UnionFind(int n) {
			parent = new int[n];
			rank = new int[n];
			for (int i = 0; i < n; i++) {
				parent[i] = i;
				rank[i] = 1;
			}
		}

		public void addEdge(int v, int u) {
			v = root(v);
			u = root(u);
			parent[u] = v;
			rank[v] += rank[u];
		}

		public boolean isConnected(int v, int u) {
			return root(v) == root(u);
		}

		public int rank(int v) {
			return rank[root(v)];
		}

		private int root(int v) {
			while (v != parent[v]) {
				parent[v] = parent[parent[v]];
				v = parent[v];
			}

			return v;
		}

	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		sc.nextLine();

		int[][] e = new int[n - 1][3];

		//int[] ea = new int[n];
		//int[] eb = new int[n];
		//int[] ew = new int[n];
		for (int i = 0; i < n - 1; i++) {
			e[i][0] = sc.nextInt();
			e[i][1] = sc.nextInt();
			e[i][2] = sc.nextInt();
		}

		Arrays.sort(e, Comparator.comparingInt(a -> a[2]));

		int[][] qs = new int[m][2];
		for (int i = 0; i < m; i++) {
			qs[i][0] = i;
			qs[i][1] = sc.nextInt();
		}

		Arrays.sort(qs, Comparator.comparing(a -> a[1]));
		long[] answers = new long[qs.length];

		UnionFind uf = new UnionFind(n + 1);

		//for (int i = 0; i < e.length; i++) {
		//	System.out.print(e[i][0]);
		//	System.out.print(" ");
		//	System.out.print(e[i][1]);
		//	System.out.print(" ");
		//	System.out.println(e[i][2]);
		//}
		//System.out.println();

		long counter = 0;
		int ei = 0;
		for (int i = 0; i < qs.length; i++) {
			int w = qs[i][1];
			for (; ei < e.length && e[ei][2] <= w; ei++) {
				counter -= (long) uf.rank(e[ei][0]) * (long) (uf.rank(e[ei][0]) - 1) / 2L;
				counter -= (long) uf.rank(e[ei][1]) * (long) (uf.rank(e[ei][1]) - 1) / 2L;
				uf.addEdge(e[ei][0], e[ei][1]);
				counter += (long) uf.rank(e[ei][1]) * (long) (uf.rank(e[ei][1]) - 1) / 2L;
			}

			answers[qs[i][0]] = counter;
		}

		for (int i = 0; i < answers.length; i++) {
			System.out.print(answers[i]);
			System.out.print(" ");
		}

		System.out.println();
	}
}