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 sys
sys.setrecursionlimit(10**9)
def find(a):
    if par[a]==a:
        return a
    par[a]=find(par[a])
    return par[a]
n,m=map(int,input().split())
ed=[]
par=[i for i in range(n)]
size=[1 for i in range(n)]
for _ in range(n-1):
    a,b,c=map(int,input().split())
    ed.append([a-1,b-1,c])
ed.sort(key=lambda x:x[2])
it=list(map(int,input().split()))
it=[[i,j,0] for j,i in enumerate(it)]
it.sort()
ind=0
tot=0
j=0
#print(it)
ss={}
for i in it[:]:
    while ind<n-1:
        if ed[ind][2]<=i[0]:
            a=find(ed[ind][0])
            b=find(ed[ind][1])
            if a!=b:
                tot+=size[a]*size[b]
              #  print(a,b,j,tot)
                if size[a]>=size[b]:

                    par[b]=a
                    size[a]+=size[b]
                    size[b]=0
                else:
                    par[a]=b
                    size[b]+=size[a]
                    size[a]=0
            ind+=1
            
        else:
            break
    it[j][2]=tot
    #ss[it[j][1]]=tot
    j+=1

it.sort(key=lambda x:x[1])
aa=[i[2] for i in it]

#for i in range(len(it)):
 #   print(ss[i],end=" ")
print(*aa)
        
    
    
        
    
