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;
const int MAXN = 200010;
long long p[MAXN], sz[MAXN];
long long res;
int find(int x) {
  if (x == p[x]) return x;
  return p[x] = find(p[x]);
}
void join(int x, int y) {
  x = find(x);
  y = find(y);
  if (x != y) {
    res += sz[x] * sz[y];
    sz[x] += sz[y];
    p[y] = x;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) sz[i] = 1, p[i] = i;
  vector<pair<long long, pair<int, int>>> v(n - 1);
  for (int i = 0, ThxDem = n - 1; i < ThxDem; ++i)
    cin >> v[i].second.first >> v[i].second.second >> v[i].first;
  for (int i = 0, ThxDem = n - 1; i < ThxDem; ++i)
    v[i].second.first--, v[i].second.second--;
  vector<pair<int, int>> qs(m);
  vector<long long> ans(m);
  for (int i = 0, ThxDem = m; i < ThxDem; ++i)
    cin >> qs[i].first, qs[i].second = i;
  sort(v.begin(), v.end());
  sort(qs.begin(), qs.end());
  int pos = 0;
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) {
    while (pos < int(v.size()) && v[pos].first <= qs[i].first) {
      join(v[pos].second.second, v[pos].second.first);
      pos++;
    }
    ans[qs[i].second] = res;
  }
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) cout << ans[i] << " ";
  cout << "\n";
}