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<int> par;
vector<int> size;
void init(int n) {
  par.clear();
  size.clear();
  par.resize(n + 1);
  size.resize(n + 1, 1);
  for (int i = 0; i <= n; i++) par[i] = i;
}
int root(int x) { return x == par[x] ? x : par[x] = root(par[x]); }
void Union(int x, int y) {
  x = root(x);
  y = root(y);
  if (size[x] < size[y]) swap(x, y);
  par[y] = x;
  size[x] += size[y];
}
int main() {
  int n, m;
  cin >> n >> m;
  init(n);
  vector<pair<int, pair<int, int> > > edges;
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    edges.push_back(make_pair(w, make_pair(u, v)));
  }
  sort(edges.begin(), edges.end());
  vector<pair<int, int> > queries;
  for (int i = 0; i < m; i++) {
    int x;
    cin >> x;
    queries.push_back(make_pair(x, i));
  }
  sort(queries.begin(), queries.end());
  vector<long long int> res(queries.size(), 0);
  int k = 0;
  long long int ans = 0;
  for (int i = 0; i < edges.size(); ++i) {
    int u = edges[i].second.first, v = edges[i].second.second,
        w = edges[i].first;
    while (k < queries.size() && queries[k].first < w)
      res[queries[k++].second] = ans;
    ans += (long long)size[root(u)] * (long long)size[root(v)];
    Union(u, v);
  }
  while (k < queries.size() && n >= 2 && queries[k].first >= edges[n - 2].first)
    res[queries[k++].second] = ans;
  for (int i = 0; i < res.size(); i++) cout << res[i] << " ";
  cout << "\n";
  return 0;
}