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 <class T, class U>
inline istream& operator>>(istream& str, pair<T, U>& p) {
  return str >> p.first >> p.second;
}
template <class T, class U>
inline ostream& operator<<(ostream& str, pair<T, U>& p) {
  return str << p.first << " " << p.second << endl;
}
template <class T>
inline istream& operator>>(istream& str, vector<T>& a) {
  for (auto& i : a) str >> i;
  return str;
}
template <class T>
inline ostream& operator<<(ostream& str, vector<T>& a) {
  for (auto& i : a) cout << i << " ";
  cout << endl;
  return str;
}
template <class T, class U>
inline ostream& operator<<(ostream& str, map<T, U>& a) {
  for (auto& i : a) cout << i.first << " " << i.second << endl;
  return str;
}
const int MAX = 200010;
std::vector<pair<int, int> > graph[MAX];
std::vector<pair<int, int> > edges, to_map(MAX);
struct DSU {
  int connected;
  int par[MAX], sz[MAX];
  DSU() {}
  DSU(int n) {
    for (int i = 1; i <= n; i++) {
      par[i] = i;
      sz[i] = 1;
    }
    connected = n;
  }
  int getPar(int k) {
    while (k != par[k]) {
      par[k] = par[par[k]];
      k = par[k];
    }
    return k;
  }
  int getSize(int k) { return sz[getPar(k)]; }
  void unite(int u, int v) {
    int par1 = getPar(u), par2 = getPar(v);
    if (par1 == par2) return;
    connected--;
    if (sz[par1] > sz[par2]) swap(par1, par2);
    sz[par2] += sz[par1];
    sz[par1] = 0;
    par[par1] = par[par2];
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    graph[u].push_back({v, w});
    edges.push_back({w, i});
    to_map[i] = make_pair(u, v);
  }
  DSU ravi(n);
  sort(edges.begin(), edges.end());
  std::vector<pair<int, int> > quer;
  long long int ans[m];
  for (int i = 0; i < m; ++i) {
    int temp;
    cin >> temp;
    quer.push_back({temp, i});
  }
  long long int final = 0;
  sort(quer.begin(), quer.end());
  int point = 0;
  for (int i = 0; i < m; i++) {
    int pre = quer[i].first;
    while (point < n - 1 and edges[point].first <= pre) {
      int idx = edges[point].second;
      final += (long long int)ravi.getSize(to_map[idx].first) *
               ravi.getSize(to_map[idx].second);
      ravi.unite(to_map[idx].first, to_map[idx].second);
      point++;
    }
    ans[quer[i].second] = final;
  }
  for (int i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}