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;
int n, k;
int v1, v2;
vector<vector<int>> g;
vector<int> w;
vector<vector<int>> dp;
vector<bool> p;
void md(int start) {
  p[start] = 1;
  vector<int> vsp;
  for (auto it : g[start])
    if (!p[it]) vsp.push_back(it), md(it);
  int size = vsp.size();
  dp[start][0] = w[start];
  for (auto it : vsp) dp[start][0] += dp[it][k];
  for (int i = 0; i != (size); i++)
    for (int j = 0; j != (k / 2); j++) {
      int sum = 0;
      for (int v = 0; v != (size); v++) sum += dp[vsp[v]][k - j - 1];
      sum += dp[vsp[i]][j] - dp[vsp[i]][k - j - 1];
      dp[start][j + 1] = max(dp[start][j + 1], sum);
    }
  for (int i = 0; i != (size); i++)
    for (int j = (k / 2); j != (k); j++) {
      int sum = 0;
      for (int v = 0; v != (size); v++) sum += dp[vsp[v]][j];
      dp[start][j + 1] = max(dp[start][j + 1], sum);
    }
  for (int i = k; i != 0; i--)
    dp[start][i - 1] = max(dp[start][i - 1], dp[start][i]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  g.resize(n), w.resize(n);
  for (int i = 0; i != (n); i++) cin >> w[i];
  for (int i = 0; i != (n - 1) && (cin >> v1 >> v2); i++)
    g[v1 - 1].push_back(v2 - 1), g[v2 - 1].push_back(v1 - 1);
  p.resize(n, 0);
  dp.resize(n, vector<int>(k + 1, 0));
  md(0);
  cout << dp[0][0];
  return 0;
}