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>
#pragma GCC optimize("Ofast")
using namespace std;
long long dsu[2000001];
long long cnt[2000001];
long long dsu_find(long long x) {
  if (x == dsu[x]) return x;
  return dsu[x] = dsu_find(dsu[x]);
}
long long ans[200001];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long> > > v;
  for (long long i = 1; i < n; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    v.push_back({c, {a, b}});
  }
  for (long long i = 1; i <= n; i++) {
    cnt[i] = 1ll;
    dsu[i] = i;
  }
  sort(v.begin(), v.end());
  vector<pair<long long, long long> > q(m);
  for (long long i = 1; i <= m; i++) {
    cin >> q[i - 1].first, q[i - 1].second = i;
  }
  sort(q.begin(), q.end());
  long long j = -1;
  long long res = 0ll;
  for (auto x : q) {
    long long i = x.first;
    long long pos = x.second;
    while (j + 1 < v.size() && v[j + 1].first <= i) {
      j++;
      long long a = v[j].second.first;
      long long b = v[j].second.second;
      res += cnt[dsu_find(a)] * cnt[dsu_find(b)] * 1ll;
      cnt[dsu_find(a)] += cnt[dsu_find(b)];
      dsu[dsu_find(b)] = dsu_find(dsu_find(a));
    }
    ans[pos] = res;
  }
  for (long long i = 1; i <= m; i++) cout << ans[i] << " ";
  return 0;
}