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>
from collections import defaultdict as ddic, deque, Counter
from itertools import combinations, permutations, product
import bisect, heapq

import sys; sys.setrecursionlimit(10**7)

class DSU:
    def __init__(self, N):
        self.par = list(range(N + 1))
        self.rnk = [0] * (N + 1)
        self.sz = [1] * (N + 1)
        self.s2 = N

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

    def union(self, x, y):
        xr, yr = self.find(x), self.find(y)
        if xr == yr: return
        if self.rnk[xr] < self.rnk[yr]:
            xr, yr = yr, xr
        if self.rnk[xr] == self.rnk[yr]:
            self.rnk[xr] += 1

        self.par[yr] = xr
        xs = self.sz[xr]
        ys = self.sz[yr]
        self.sz[xr] += self.sz[yr]
        self.s2 += 2 * xs * ys

    def size(self, x):
        return self.sz[self.find(x)]


rr = raw_input
rri = lambda: int(raw_input())
rrm = lambda: map(int, raw_input().split())

def solve(N, M, edges, queries):
    queries = [(q, i) for i, q in enumerate(queries)]
    queries.sort()
    edges.sort(key = lambda (u,v,w): w)
    eix = 0
    ans = [None] * M
    dsu = DSU(N)

    for q, i in queries:
        # edges <= q are allowed
        while eix < len(edges) and edges[eix][2] <= q:
            dsu.union(edges[eix][0], edges[eix][1])
            eix += 1

        bns = (dsu.s2 - N) / 2
        ans[i] = bns
    return ans
        
        


N, M = rrm()
edges = []
for _ in xrange(N-1):
    u, v, w = rrm()
    u -= 1
    v -= 1
    edges.append((u,v,w))

queries = rrm()

ans = solve(N, M, edges, queries)
print " ".join(map(str, ans))
