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, k = map(int, input().split())

a = list(map(int, input().split()))
g = [[] for i in range(n)]
for i in range(n - 1):
	v, u = map(int, input().split())
	v -= 1
	u -= 1
	g[v].append(u)
	g[u].append(v)

dp = [[] for i in range(n)]
d = [1 for i in range(n)]

def dfs(v, p = -1):
	dp[v].append(a[v])
	for u in g[v]:
		if u == p:
			continue
		dfs(u, v)
		tmp = [-10**18 for i in range(max(d[v], d[u] + 1))]
		for i in range(d[v]):
			for j in range(max(0, k - i), d[u]):
				tmp[min(i, j + 1)] = max(tmp[min(i, j + 1)], dp[v][i] + dp[u][j])
		for i in range(d[u]):
			tmp[i + 1] = max(tmp[i + 1], dp[u][i])
		for i in range(d[v]):
			dp[v][i] = max(dp[v][i], tmp[i])
		dp[v] += tmp[d[v]:]
		d[v] = max(d[v], d[u] + 1)
		for i in range(d[v] - 1, 0, -1):
			dp[v][i - 1] = max(dp[v][i - 1], dp[v][i])

dfs(0)
print(max(dp[0]))