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;
using namespace chrono;
template <class T1, class T2>
istream &operator>>(istream &in, pair<T1, T2> &P) {
  in >> P.first >> P.second;
  return in;
}
template <class T1, class T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &P) {
  out << "(" << P.first << ", " << P.second << ")";
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &arr) {
  for (auto &x : arr) in >> x;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &arr) {
  for (auto &x : arr) out << x << ' ';
  cout << "\n";
  return out;
}
template <class T>
istream &operator>>(istream &in, deque<T> &arr) {
  for (auto &x : arr) in >> x;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const deque<T> &arr) {
  for (auto &x : arr) out << x << ' ';
  cout << "\n";
  return out;
}
struct disjoint {
  int n;
  vector<int> parent;
  vector<long long> size;
  disjoint(int _n) : n(_n), parent(n), size(n, 1) {
    iota(parent.begin(), parent.end(), 0);
  }
  int insert() {
    parent.push_back(parent.size());
    return parent.size() - 1;
  }
  int root(int u) { return parent[u] == u ? u : (parent[u] = root(parent[u])); }
  long long solve(int u) { return size[root(u)] * (size[root(u)] - 1) / 2; }
  void merge(int u, int v) {
    if (root(u) != root(v)) size[root(u)] += size[root(v)];
    parent[root(v)] = root(u);
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<tuple<int, int, int>> e(n - 1);
  for (auto &[w, u, v] : e) {
    cin >> u >> v >> w, u--, v--;
  }
  vector<pair<int, int>> query(m);
  for (int i = 0; i < m; i++) {
    cin >> query[i].first, query[i].second = i;
  }
  sort(e.begin(), e.end(), greater<tuple<int, int, int>>{}),
      sort(query.begin(), query.end(), greater<pair<int, int>>{});
  long long cnt = 0;
  vector<long long> res(m);
  disjoint dsu(n);
  while (!query.empty()) {
    auto q = query.back();
    query.pop_back();
    while (!e.empty() && get<0>(e.back()) <= q.first) {
      auto [w, u, v] = e.back();
      e.pop_back();
      cnt -= dsu.solve(u) + dsu.solve(v);
      dsu.merge(u, v);
      cnt += dsu.solve(u);
    }
    res[q.second] = cnt;
  }
  for (auto x : res) cout << x << " ";
  return 0;
}