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>
# -*- coding: utf-8 -*-

import sys
from itertools import accumulate

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10 ** 9 + 7

class UnionFind:
    def __init__(self, n):
        self.n = n
        self.par = [i for i in range(n+1)]
        self.rank = [0] * (n+1)
        self.size = [1] * (n+1)
        self.tree = [True] * (n+1)

    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            self.tree[x] = False
            return
        if not self.tree[x] or not self.tree[y]:
            self.tree[x] = self.tree[y] = False

        if self.rank[x] < self.rank[y]:
            self.par[x] = y
            self.size[y] += self.size[x]
        else:
            self.par[y] = x
            self.size[x] += self.size[y]
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

    def is_same(self, x, y):
        return self.find(x) == self.find(y)

    def get_size(self, x=None):
        if x is not None:
            return self.size[self.find(x)]
        else:
            res = set()
            for i in range(self.n+1):
                res.add(self.find(i))
            return len(res) - 1
    
    def is_tree(self, x):
        return self.tree[self.find(x)]

N, M = MAP()
edges = []
for i in range(N-1):
    a, b, c = MAP()
    a -= 1; b -= 1
    edges.append((c, a, b))

if N == 1:
    LIST()
    ans = [0] * M
    print(*ans)
    exit()

edges.sort()
A = [0] * (2*10**5+7)
uf = UnionFind(N)
prevc = edges[0][0]
for c, a, b in edges:
    if prevc != c:
        A[c] = A[prevc]
    sz1 = uf.get_size(a)
    A[c] -= sz1 * (sz1-1) // 2
    sz2 = uf.get_size(b)
    A[c] -= sz2 * (sz2-1) // 2
    uf.union(a, b)
    sz = sz1 + sz2
    A[c] += sz * (sz-1) // 2
    prevc = c

A = list(accumulate(A, max))
Q = LIST()
ans = [0] * M
for i, q in enumerate(Q):
    ans[i] = A[q]
print(*ans)
