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;
class DisjointSet {
 public:
  DisjointSet(int size)
      : fathers_(size, -1), sizes_(size, 1), total_pairs_(0) {}
  void Unite(int a, int b);
  int64_t TotalPairs() const { return total_pairs_; }

 private:
  int Root(int node);
  vector<int> fathers_;
  vector<int> sizes_;
  int64_t total_pairs_;
};
int DisjointSet::Root(int node) {
  if (fathers_[node] == -1) {
    return node;
  }
  return fathers_[node] = Root(fathers_[node]);
}
int64_t CountOrderedPairs(int size) { return 1LL * size * (size - 1) / 2; }
void DisjointSet::Unite(int a, int b) {
  int root_a = Root(a);
  int root_b = Root(b);
  if (root_a != root_b) {
    fathers_[root_b] = root_a;
    total_pairs_ -= CountOrderedPairs(sizes_[root_b]);
    total_pairs_ -= CountOrderedPairs(sizes_[root_a]);
    sizes_[root_a] += sizes_[root_b];
    sizes_[root_b] = 0;
    total_pairs_ += CountOrderedPairs(sizes_[root_a]);
  }
}
struct Edge {
  int a, b;
  int cost;
};
int FindIndex(const vector<int> &vec, int max_val) {
  int index = -1;
  int power = (1 << 25);
  while (power > 0) {
    int next_index = index + power;
    power >>= 1;
    if (next_index < (int)vec.size() && vec[next_index] <= max_val) {
      index = next_index;
    }
  }
  return index;
}
int main() {
  int nodes, queries;
  cin >> nodes >> queries;
  vector<Edge> edges(nodes - 1);
  for (auto &e : edges) {
    cin >> e.a >> e.b >> e.cost;
    e.a -= 1;
    e.b -= 1;
  }
  sort(edges.begin(), edges.end(),
       [](const Edge &a, const Edge &b) { return a.cost < b.cost; });
  DisjointSet set(nodes);
  vector<int> costs;
  vector<int64_t> answers;
  for (size_t i = 0; i < edges.size(); i += 1) {
    if (i == 0 || edges[i].cost != edges[i - 1].cost) {
      costs.push_back(edges[i].cost);
      answers.push_back(0);
    }
    set.Unite(edges[i].a, edges[i].b);
    answers.back() = set.TotalPairs();
  }
  for (int i = 0; i < queries; i += 1) {
    int max_cost;
    cin >> max_cost;
    int index = FindIndex(costs, max_cost);
    cout << (index < 0 ? 0 : answers[index]) << " ";
  }
  cout << "\n";
  return 0;
}