Problem: 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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long inf = 0x7f7f7f7f;
const long long mod = 1e9 + 7;
const int N = 1000005;
int n, k, a[N], h[N], nxt[N], adj[N], t, nw, u, v, d[N], ans;
vector<int> g[N];
void add() {
  nxt[++t] = h[u], h[u] = t, adj[t] = v, nxt[++t] = h[v], h[v] = t, adj[t] = u;
}
void dfs(int x, int fa) {
  g[d[x] = d[fa] + 1].push_back(x);
  for (int i = h[x], j; i; i = nxt[i])
    if ((j = adj[i]) != fa) dfs(j, x);
}
void Dfs(int x, int fa, int dep) {
  a[x] -= nw;
  if (dep == k) return;
  for (int i = h[x], j; i; i = nxt[i])
    if ((j = adj[i]) != fa) Dfs(j, x, dep + 1);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) cin >> u, cin >> v, add();
  dfs(1, 0);
  for (int i = n; i; --i)
    for (int j = 0, sz = g[i].size(), x; j < sz; ++j)
      if ((x = a[g[i][j]]) > 0) nw = x, ans += x, Dfs(g[i][j], 0, 0);
  printf("%d", ans);
  return 0;
}