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;
template <typename T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
class UnionFind {
 private:
  int n;
  vector<int> parent, r;
  int _find(int x) {
    assert(x < n);
    if (parent[x] == x)
      return x;
    else
      return parent[x] = _find(parent[x]);
  }
  void _unite(int x, int y) {
    x = _find(x);
    y = _find(y);
    if (x == y) return;
    if (r[x] < r[y]) {
      parent[x] = y;
      r[y] += r[x];
    } else {
      parent[y] = x;
      r[x] += r[y];
    }
  }

 public:
  int find(int x) { return _find(x); }
  void unite(int x, int y) { _unite(x, y); }
  int count(int x) { return r[_find(x)]; }
  int operator[](int i) { return _find(i); }
  UnionFind() : n(0) {}
  explicit UnionFind(size_t _n) : n(_n) {
    parent = vector<int>(n);
    r = vector<int>(n);
    for (int i = 0; i < n; i++) {
      parent[i] = i;
      r[i] = 1;
    }
  }
};
class Task {
 public:
  void solve(istream& in, ostream& out) {
    int N, M;
    in >> N >> M;
    const int MAX_W = (int)2e5;
    vector<vector<pair<int, int>>> E(MAX_W + 1);
    for (int i = (0); i < (N - 1); ++i) {
      int u, v, w;
      in >> u >> v >> w;
      u--, v--;
      E[w].push_back({u, v});
    }
    vector<long long> ans(MAX_W + 1, 0);
    UnionFind uf(N);
    for (int i = (1); i < (MAX_W + 1); ++i) {
      ans[i] = ans[i - 1];
      for (auto p : E[i]) {
        if (uf[p.first] != uf[p.second]) {
          ans[i] += 1ll * uf.count(p.first) * uf.count(p.second);
          uf.unite(p.first, p.second);
        }
      }
    }
    while (M--) {
      int q;
      in >> q;
      out << ans[q] << ' ';
    }
    out << '\n';
  }
};
int main() {
  istream& in(cin);
  ostream& out(cout);
  ios_base::sync_with_stdio(false);
  in.tie(nullptr), out.tie(nullptr);
  Task solver;
  solver.solve(in, out);
  return 0;
}