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;
long long find(vector<long long> &parent, vector<long long> &rank,
               long long x) {
  while (parent[x] != x) {
    parent[x] = parent[parent[x]];
    x = parent[x];
  }
  return x;
}
void union1(vector<long long> &parent, vector<long long> &rank, long long x,
            long long y) {
  long long p1 = find(parent, rank, x);
  long long p2 = find(parent, rank, y);
  if (p1 == p2) return;
  if (rank[p1] > rank[p2]) {
    parent[p2] = p1;
    rank[p1] += rank[p2];
  } else {
    parent[p1] = p2;
    rank[p2] += rank[p1];
  }
}
int main() {
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long> > > edges(n - 1);
  vector<pair<long long, long long> > q(m);
  vector<long long> ans(m);
  for (long long i = 0; i < n - 1; i++) {
    cin >> edges[i].second.first >> edges[i].second.second >> edges[i].first;
    edges[i].second.first--;
    edges[i].second.second--;
  }
  long long tmp;
  for (long long i = 0; i < m; i++) {
    cin >> tmp;
    q[i] = {tmp, i};
  }
  sort(edges.begin(), edges.end());
  sort(q.begin(), q.end());
  vector<long long> parent(n, 0);
  vector<long long> rank(n, 1);
  iota(parent.begin(), parent.end(), 0);
  long long j = 0;
  long long counter = 0;
  for (long long i = 0; i < q.size(); i++) {
    long long tq = q[i].first;
    while (j < edges.size() && edges[j].first <= tq) {
      long long p1 = find(parent, rank, edges[j].second.first);
      long long p2 = find(parent, rank, edges[j].second.second);
      counter -= (rank[p1] * (rank[p1] - 1)) / 2;
      counter -= (rank[p2] * (rank[p2] - 1)) / 2;
      union1(parent, rank, p1, p2);
      counter += (rank[p1] > rank[p2]) ? (rank[p1] * (rank[p1] - 1)) / 2
                                       : (rank[p2] * (rank[p2] - 1)) / 2;
      ++j;
    }
    ans[q[i].second] = counter;
  }
  cout << ans[0];
  for (long long i = 1; i < m; i++) {
    cout << " " << ans[i];
  }
  cout << endl;
  return 0;
}