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 sys import setrecursionlimit as SRL, stdin

SRL(10 ** 7)
rd = stdin.readline
rrd = lambda: map(int, rd().strip().split())

fa = [i for i in range(200005)]
s = [1] * 200005


def find(x):
    t = []
    while fa[x] != x:
        t.append(x)
        x = fa[x]
    for i in t:
        fa[i] = x
    return fa[x]


ans = [0] * 200005
n, q = rrd()

w = []
for i in range(n - 1):
    x, y, z = rrd()
    w.append([z, x, y])

w.sort(key=lambda x: x[0])
for x in w:
    u = find(x[1])
    v = find(x[2])

    ans[x[0]] += s[u] * s[v]
    fa[u] = v
    s[v] += s[u]

for i in range(1, 200001):
    ans[i] += ans[i - 1]

q = list(rrd())

for x in q:
    print(ans[x], end=' ')
