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;
class UnionFind {
 private:
  vector<int> rank;
  vector<int> p;
  vector<int> sz;

 public:
  UnionFind(int n) { reset(n); }
  void reset(int n) {
    rank.assign(n, 0);
    sz.assign(n, 1);
    p.assign(n, 0);
    for (int i = 0; i < n; ++i) p[i] = i;
  }
  int find(int i) {
    if (i == p[i]) return i;
    return (p[i] = find(p[i]));
  }
  void merge(int i, int j) {
    i = find(i);
    j = find(j);
    if (i != j) {
      if (rank[i] > rank[j]) {
        p[j] = i;
        sz[i] += sz[j];
      } else {
        p[i] = j;
        sz[j] += sz[i];
        if (rank[i] == rank[j]) rank[j]++;
      }
    }
  }
  int size(int i) { return sz[find(i)]; }
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, m;
  cin >> n >> m;
  UnionFind uf(n);
  vector<pair<long long, pair<long long, long long> > > edges;
  for (long long i = 0; i < n - 1; ++i) {
    long long u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    edges.push_back({w, {u, v}});
  }
  sort(edges.begin(), edges.end());
  map<long long, long long> dp;
  long long calc = 0;
  for (long long i = 0; i < n - 1; ++i) {
    long long u = edges[i].second.first, v = edges[i].second.second;
    long long szu = uf.size(u), szv = uf.size(v);
    calc = calc + szu * szv;
    uf.merge(u, v);
    dp[edges[i].first] = calc;
  }
  for (long long i = 0; i < m; ++i) {
    long long q;
    cin >> q;
    auto x = dp.lower_bound(q);
    if (x == dp.end()) {
      cout << n * (n - 1) / 2 << ' ';
    } else {
      if (x == dp.begin()) {
        if (x->first > q) {
          cout << "0 ";
        } else {
          cout << x->second << ' ';
        }
      } else {
        if (x->first > q) {
          x--;
        }
        cout << x->second << ' ';
      }
    }
  }
  cout << endl;
  return 0;
}