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>
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
import sys
from sys import stdin, stdout
from collections import defaultdict
from collections import deque
import math
import copy

#T = int(input())
#N = int(input())
#s1 = input()
#s2 = input()
N,E = [int(x) for x in stdin.readline().split()]
#arr = [int(x) for x in stdin.readline().split()]

s = 0
size = [1]*N

f = [x for x in range(N)]

def getf(v):
    stack = []
    while f[v] != v:
       stack.append(v)
       v = f[v]
    for idx in stack:
        f[idx] = v
    return v

def merge(f, v, u):
    global s
    t1 = getf(v)
    t2 = getf(u)
    if t1 != t2:
        A = size[t1]
        B = size[t2]
        s -= A*(A-1)//2
        s -= B*(B-1)//2
        size[t1] = size[t1] + size[t2]
        C = size[t1]
        s += C*(C-1)//2
        f[t2] = t1


w_edge = {}
for i in range(N-1):
    u,v,w = [int(x) for x in stdin.readline().split()]
    if u>v:
        u,v = v,u

    if w not in w_edge:
        w_edge[w] = []

    w_edge[w].append((u,v))

q = [0]*200000
for i in range(200000):
    if i+1 in w_edge:
        for e in w_edge[i+1]:
            u,v = e
            merge(f,u-1,v-1)
        q[i] = s
    else:
        if i!=0:
            q[i] = q[i-1]
        else:
            q[i] = 0

arr = [int(x) for x in stdin.readline().split()]
ans = [0]*E
for i in range(E):
    ans[i] = q[arr[i]-1]

print(*ans)
