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>
#pragma GCC optimize("O2")
using namespace std;
vector<long long> arr;
vector<vector<int>> G;
int n, k;
vector<int> dfs(int x, int p) {
  vector<vector<int>> tmp;
  for (int i : G[x]) {
    if (i != p) {
      tmp.push_back(dfs(i, x));
    }
  }
  vector<int> ret(k + 1, 0);
  ret[0] = arr[x];
  for (int i = 0; i < tmp.size(); i++) ret[0] += tmp[i][k];
  for (int i = 1; i < k; i++) {
    if (i + i < k) {
      for (int j = 0; j < tmp.size(); j++)
        ret[i] = max(ret[i], tmp[j][i] - tmp[j][k - i]);
    }
    for (int j = 0; j < tmp.size(); j++) ret[i] += tmp[j][max(i, k - i)];
  }
  for (int i = k; i; i--) ret[i] = ret[i - 1];
  ret[0] = 0;
  for (int i = k - 1; i; i--) ret[i] = max(ret[i], ret[i + 1]);
  return ret;
}
int main() {
  cin.tie(0), ios::sync_with_stdio(0);
  cin >> n >> k;
  ++k;
  arr.resize(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  G.resize(n);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    G[a - 1].push_back(b - 1);
    G[b - 1].push_back(a - 1);
  }
  cout << dfs(0, 0)[1] << "\n";
  return 0;
}