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 ll = long long;
using ull = unsigned long long;
ll res = 0;
vector<int> parent, tot;
int getp(int u) {
  if (parent[u] == u) return u;
  return parent[u] = getp(parent[u]);
}
ll bino(int u) { return (1LL * u * 1LL * (u - 1)) / 2; }
void merge(int u, int v) {
  u = getp(u);
  v = getp(v);
  res -= bino(tot[u]);
  res -= bino(tot[v]);
  if (u < v) swap(u, v);
  tot[u] += tot[v];
  res += bino(tot[u]);
  parent[v] = u;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  parent = tot = vector<int>(n, 1);
  iota(parent.begin(), parent.end(), 0);
  vector<pair<int, pair<int, int>>> e(n - 1);
  for (int i = 0; i < n - 1; i++) {
    cin >> e[i].second.first >> e[i].second.second >> e[i].first;
    e[i].second.first--;
    e[i].second.second--;
  }
  vector<pair<int, int>> q(m);
  vector<ll> ans(m);
  for (int i = 0; i < m; i++) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(e.begin(), e.end());
  sort(q.begin(), q.end());
  int pos = 0;
  for (int i = 0; i < m; i++) {
    while (pos < n - 1 and e[pos].first <= q[i].first) {
      merge(e[pos].second.first, e[pos].second.second);
      pos++;
    }
    ans[q[i].second] = res;
  }
  for (int i = 0; i < m; i++) cout << ans[i] << " ";
  cout << endl;
}