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;
using T = long long int;
T par[200005];
T root(T v) { return (par[v] < 0) ? v : par[v] = root(par[v]); }
void merge(T x, T y) {
  if ((x = root(x)) == (y = root(y))) return;
  if (par[x] > par[y]) swap(x, y);
  par[x] += par[y];
  par[y] = x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  T TESTS = 1;
  while (TESTS--) {
    T n, m, i, j, k;
    cin >> n >> m;
    if (n == 1) {
      for (i = 0; i < m; i++) cin >> j;
      for (i = 0; i < m; i++) cout << 0 << " ";
      return 0;
    }
    memset(par, -1, sizeof par);
    vector<pair<T, pair<T, T> > > edges(n - 1);
    for (i = 0; i < n - 1; i++) {
      T x, y, wt;
      cin >> x >> y >> wt;
      edges[i] = make_pair(wt, make_pair(x, y));
    }
    sort(edges.begin(), edges.end());
    map<T, T> ans;
    for (i = 0; i < n - 1; i++) {
      T x = edges[i].second.first, y = edges[i].second.second,
        wt = edges[i].first;
      ans[wt] += (par[root(x)] * par[root(y)]);
      merge(x, y);
    }
    vector<T> weights;
    for (auto i : ans) weights.push_back(i.first);
    vector<T> pref(weights.size(), 0);
    pref[0] = ans[weights[0]];
    for (i = 1; i < weights.size(); i++)
      pref[i] = pref[i - 1] + ans[weights[i]];
    while (m--) {
      T wgt;
      cin >> wgt;
      T idx =
          upper_bound(weights.begin(), weights.end(), wgt) - weights.begin();
      idx--;
      if (idx >= 0)
        cout << pref[idx] << " ";
      else
        cout << "0 ";
    }
  }
  return 0;
}