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 mod = 1000000007;
vector<long long> parent(200005), Size(200005, 1);
long long answer = 0;
long long find_set(long long n) {
  if (parent[n] == n)
    return n;
  else
    return find_set(parent[n]);
}
void add_edge(long long a, long long b) {
  long long A = find_set(a);
  long long B = find_set(b);
  if (Size[A] >= Size[B]) {
    answer -= ((Size[A] * (Size[A] - 1)) / 2);
    answer -= ((Size[B] * (Size[B] - 1)) / 2);
    Size[A] += Size[B];
    Size[B] = 0;
    parent[B] = A;
    answer += ((Size[A] * (Size[A] - 1)) / 2);
  } else {
    answer -= ((Size[B] * (Size[B] - 1)) / 2);
    answer -= ((Size[A] * (Size[A] - 1)) / 2);
    Size[B] += Size[A];
    Size[A] = 0;
    parent[A] = B;
    answer += ((Size[B] * (Size[B] - 1)) / 2);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long N, Q, p, q, r;
  cin >> N >> Q;
  if (N == 1) {
    for (long long i = 1; i <= Q; i++) cout << "0 ";
    return 0;
  }
  for (long long i = 1; i <= N; i++) parent[i] = i;
  vector<pair<long long, pair<long long, long long>>> G;
  for (long long i = 1; i < N; i++) {
    cin >> p >> q >> r;
    G.push_back({r, {p, q}});
  }
  sort(G.begin(), G.end());
  vector<long long> final_ans(N + 1);
  long long ptr = 0;
  for (long long i = 1; i <= N;) {
    if (G[ptr].first == i) {
      add_edge(G[ptr].second.first, G[ptr].second.second);
      ptr++;
    } else {
      final_ans[i] = answer;
      i++;
    }
  }
  while (Q--) {
    long long k;
    cin >> k;
    cout << final_ans[k] << ' ';
  }
  return 0;
}