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 dsu {
 public:
  vector<int> p, sz;
  int n;
  long long ans;
  dsu(int _n) : n(_n) {
    p.resize(n);
    sz = vector<int>(n, 1);
    iota(p.begin(), p.end(), 0);
    ans = 0;
  }
  inline int get(int x) { return (x == p[x] ? x : (p[x] = get(p[x]))); }
  inline bool unite(int x, int y) {
    x = get(x);
    y = get(y);
    if (x != y) {
      ans += 1LL * sz[y] * sz[x];
      sz[y] += sz[x];
      p[x] = y;
      return true;
    }
    return false;
  }
  inline long long get_ans() { return ans; }
};
int main() {
  ios::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  vector<tuple<int, int, int> > e(n - 1);
  for (int i = 0; i < n - 1; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    e[i] = make_tuple(c, a, b);
  }
  sort(e.begin(), e.end());
  vector<pair<int, int> > q(m);
  for (int i = 0; i < m; ++i) {
    cin >> q[i].first;
    q[i].second = i;
  }
  vector<long long> ans(m);
  sort(q.begin(), q.end());
  dsu ds(n);
  for (int i = 0, j = 0; i < (int)q.size(); i++) {
    while (j < (int)e.size() && get<0>(e[j]) <= q[i].first) {
      ds.unite(get<1>(e[j]), get<2>(e[j]));
      j++;
    }
    ans[q[i].second] = ds.get_ans();
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << ' ';
  return 0;
}