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 Forest {
 private:
  vector<int> card;
  vector<int> height;
  vector<int> father;

 public:
  Forest(int n) {
    card.resize(n + 1, 1);
    height.resize(n + 1);
    father.resize(n + 1);
  }
  int getCard(int root) { return card[root]; }
  int find(int x) {
    int root = x;
    while (father[root]) root = father[root];
    int aux;
    while (father[x]) {
      aux = father[x];
      father[x] = root;
      x = aux;
    }
    return root;
  }
  void unite(int rootX, int rootY) {
    if (height[rootX] < height[rootY]) {
      father[rootX] = rootY;
      card[rootY] += card[rootX];
    } else {
      father[rootY] = rootX;
      card[rootX] += card[rootY];
      if (height[rootX] == height[rootY]) height[rootX]++;
    }
  }
};
struct Edge {
  int x, y, cost;
  Edge(int x, int y, int cost) : x(x), y(y), cost(cost) {}
};
inline bool operator<(Edge x, Edge y) { return x.cost > y.cost; }
class Graph {
 private:
  int n;
  vector<Edge> edges;

 public:
  Graph(int n) { this->n = n; }
  void addEdge(int x, int y, int cost) { edges.emplace_back(x, y, cost); }
  vector<int64_t> kruskal(vector<int>& v) {
    Forest forest(n);
    make_heap(edges.begin(), edges.end());
    auto cnt = [](int x) { return 1LL * x * (x - 1) / 2; };
    int ind = 0;
    vector<int64_t> sol(v.size());
    for (int i = 1; i < n;) {
      Edge edge = edges[0];
      pop_heap(edges.begin(), edges.end());
      edges.pop_back();
      int rootX = forest.find(edge.x);
      int rootY = forest.find(edge.y);
      if (rootX != rootY) {
        int cardX = forest.getCard(rootX);
        int cardY = forest.getCard(rootY);
        int64_t upd = cnt(cardX + cardY) - cnt(cardX) - cnt(cardY);
        if (edge.cost <= v[ind])
          sol[ind] += upd;
        else {
          for (ind++; ind < (int)v.size() && v[ind] < edge.cost; ind++)
            sol[ind] = sol[ind - 1];
          if (ind == (int)v.size()) return sol;
          sol[ind] = sol[ind - 1] + upd;
        }
        i++;
        forest.unite(rootX, rootY);
      }
    }
    for (int i = ind + 1; i < (int)v.size(); i++) sol[i] = sol[ind];
    return sol;
  }
};
int main() {
  int n, q;
  cin >> n >> q;
  Graph graph(n);
  for (int i = 1; i < n; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    graph.addEdge(x, y, z);
  }
  vector<int> v(q);
  vector<pair<int64_t, int>> w(q);
  for (int i = 0; i < q; i++) {
    cin >> v[i];
    w[i].first = v[i];
    w[i].second = i;
  }
  sort(v.begin(), v.end());
  sort(w.begin(), w.end());
  vector<int64_t> sol = graph.kruskal(v);
  for (int i = 0; i < q; i++) w[i].first = sol[i];
  sort(w.begin(), w.end(), [](pair<int64_t, int> x, pair<int64_t, int> y) {
    return x.second < y.second;
  });
  for (int i = 0; i < q; i++) cout << w[i].first << ' ';
  cout << '\n';
  return 0;
}