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;
vector<pair<long long, long long> > g[200100];
long long pr[200100];
long long cnt[200100];
long long ans = 0;
long long get(long long v) {
  while (pr[v] != v) {
    pr[v] = pr[pr[v]];
    v = pr[v];
  }
  return v;
}
void merge(long long a, long long b) {
  long long f = get(a), s = get(b);
  if (f != s) {
    ans -= (cnt[s] * (cnt[s] - 1) / 2);
    ans -= (cnt[f] * (cnt[f] - 1) / 2);
    pr[s] = f;
    cnt[f] += cnt[s];
    ans += (cnt[f] * (cnt[f] - 1) / 2);
  }
}
long long ans_a[200100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    pr[i] = i;
    cnt[i] = 1;
  }
  vector<pair<long long, pair<long long, long long> > > v;
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b, w;
    cin >> a >> b >> w;
    g[a].push_back(make_pair(b, w));
    g[b].push_back(make_pair(a, w));
    v.push_back(make_pair(w, make_pair(a, b)));
  }
  vector<pair<long long, long long> > q;
  for (long long i = 1; i <= m; i++) {
    long long x;
    cin >> x;
    q.push_back(make_pair(x, i));
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  sort(q.begin(), q.end());
  for (long long i = 0; i < q.size(); i++) {
    long long x = q[i].first;
    while (v.size() && v.back().first <= x) {
      long long f = v.back().second.first;
      long long s = v.back().second.second;
      merge(f, s);
      v.pop_back();
    }
    ans_a[q[i].second] = ans;
  }
  for (long long i = 1; i <= m; i++) {
    cout << ans_a[i] << " ";
  }
  cout << endl;
  return 0;
}