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>
n, m = map(int, input().split())

edges = []
dsu = [i for i in range(n + 1)]
ans = [0 for i in range(2 * 10 ** 5 + 1)]
gang_num = [1 for i in range(n + 1)]
for i in range(n - 1):
    u, v, w = map(int, input().split())
    edges.append((w, u, v))

edges.sort()


def cal_num(n):
    return (n*(n - 1)) // 2


def union(u, v):
    first, second = find(u), find(v)
    if first == second:
        return 0,0,0
    dsu[first] = second
    prev_f,prev_s = gang_num[first],gang_num[second]
    gang_num[second] += gang_num[first]
    return gang_num[second],prev_f,prev_s


def find(u):
    to_find = u
    while u != dsu[u]:
        u = dsu[u]
    dsu[to_find] = u

    return u

prev = 0
for each_edge in edges:
    w, u, v = each_edge
    big, smallf, smalls = union(u, v)
    ans[w] = prev + cal_num(big) - cal_num(smallf) - cal_num(smalls)
    prev = ans[w]
for i in range(1, len(ans)):
    if ans[i] == 0:
        ans[i] = ans[i - 1]

query = [int(i) for i in input().split()]
print(' '.join(str(ans[i]) for i in query))


