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.*;
import java.io.*;
public class Solution{
	public static Integer INT(String s){
		return Integer.parseInt(s);
	}

	public static Long LONG(String s){
		return Long.parseLong(s);
	}

	//====================================================================================================================



	

	public static void main(String args[])throws IOException{

		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));	Scanner in=new Scanner(System.in); StringBuilder out=new StringBuilder();

		
		int n=in.nextInt(), 
			m=in.nextInt();

		Integer edge[][]=new Integer[n-1][3];
		for(int i=0; i<n-1; i++) 
			for(int j=0; j<3; j++)
				edge[i][j]=in.nextInt();

		Arrays.sort(edge, (Integer a1[], Integer a2[])->{
			return a1[2]-a2[2];
		});

		DSU obj=new DSU(n);

		long q[]=new long[1000_000];
		for(int i=0; i<n-1; i++){

			if(i>0 && edge[i][2]!=edge[i-1][2]) 
				q[edge[i-1][2]]=obj.count();

			obj.insert(edge[i][0], edge[i][1]);
		}
		
		if(n!=1)
			q[edge[n-2][2]]=obj.count();

		for(int i=1; i<1000_000; i++)
			if(q[i]==0)
				q[i]=q[i-1];

		for(int i=0; i<m; i++)
			out.append(q[in.nextInt()]+" ");

		System.out.print(out);
	}
}

class DSU {

	int p[], depth[];
	HashMap<Integer, Integer> map;

	DSU(int n) {
		p=new int[1000_000];
		depth=new int[1000_000];
		for(int i=0; i<1000_000; i++) {
			p[i]=i;
			depth[i]=1;
		}
		map=new HashMap<>();
		map.put(1, n);
	}

	public int parent(int x) {
		if(p[x]==x)
			return x;
		return p[x]=parent(p[x]);
	}

	public void insert(int x, int y) {

		x=parent(x);
		y=parent(y);

		if(x==y)
			return;

		if(depth[x]<depth[y]) {
			int temp=x;
			x=y;
			y=temp;
		}

		ifEmpty(depth[x]);
		ifEmpty(depth[y]);

		p[y]=p[x];
		depth[x]+=depth[y];
		map.put(depth[x], map.getOrDefault(depth[x], 0)+1);
	}

	public long count() {

		long cnt=0;
		for(Map.Entry<Integer, Integer> pair: map.entrySet()) {
			long key=pair.getKey(),
				 value=pair.getValue();
			cnt+=(key*(key-1))/2L*value;
		}

		return cnt;
	}

	public void ifEmpty(int x) {
		map.put(x, map.get(x)-1);
		if(map.get(x)==0)
			map.remove(x);
	}
}