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>
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
using namespace std;
long long par[200005], size[200005];
long long root(long long i) {
  if (par[i] == i) return i;
  return par[i] = root(par[i]);
}
void merge(long long a, long long b) {
  long long p = root(a);
  long long q = root(b);
  if (size[p] > size[q]) swap(p, q);
  par[p] = q;
  size[q] += size[p];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, n, m, k, cnt = 0, ans = 0, t = 1;
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long> > > e;
  for (i = 0; i < n - 1; i++) {
    long long w;
    cin >> j >> k >> w;
    e.push_back(make_pair(w, make_pair(j, k)));
  }
  sort(e.begin(), e.end());
  vector<pair<long long, long long> > q;
  for (i = 0; i < m; i++) {
    cin >> j;
    q.push_back(make_pair(j, i));
  }
  sort(q.begin(), q.end());
  for (i = 0; i < 200005; i++) par[i] = i, size[i] = 1;
  set<long long> s;
  for (i = 1; i < n + 1; i++) {
    if (root(i) == i) s.insert(i);
  }
  ans = 0;
  j = 0;
  long long out[m + 1];
  for (i = 0; i < m; i++) {
    while (j < e.size() && e[j].first <= q[i].first) {
      long long v1 = e[j].second.first, v2 = e[j].second.second;
      if (root(v1) != root(v2)) {
        long long a = root(v1);
        long long b = root(v2);
        if (size[a] > size[b]) {
          s.erase(b);
        } else
          s.erase(a);
        ans -= (size[a] * (size[a] - 1)) / 2;
        ans -= (size[b] * (size[b] - 1)) / 2;
        ans += ((size[a] + size[b]) * (size[a] + size[b] - 1)) / 2;
        merge(v1, v2);
      }
      j++;
    }
    out[q[i].second] = ans;
  }
  for (long long it = 0; it < m; it++) cout << out[it] << " ";
  cout << '\n';
  return 0;
}