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 long long mod = 1000000007;
const long long infinity = 1000000000000000000;
const int inf = 1e9 + 5;
bool do_debug = false;
long long running_ans = 0;
vector<vector<int>> component;
vector<int> component_of;
int cp;
void dsu(int x, int y) {
  if (component_of[x] == -1 && component_of[y] == -1) {
    running_ans += 1;
    component.push_back({x, y});
    component_of[x] = component_of[y] = cp++;
  } else if (component_of[x] == -1) {
    component_of[x] = component_of[y];
    running_ans += component[component_of[y]].size();
    component[component_of[x]].push_back(x);
  } else if (component_of[y] == -1) {
    component_of[y] = component_of[x];
    running_ans += component[component_of[x]].size();
    component[component_of[y]].push_back(y);
  } else {
    if (component_of[x] == component_of[y]) {
      return;
    }
    int sx = component[component_of[x]].size();
    int sy = component[component_of[y]].size();
    running_ans += (1LL * sx) * sy;
    if (sx > sy) {
      swap(x, y);
    }
    for (int node : component[component_of[x]]) {
      if (node != x) {
        component_of[node] = component_of[y];
        component[component_of[y]].push_back(node);
      }
    }
    component[component_of[y]].push_back(x);
    component[component_of[x]].clear();
    component_of[x] = component_of[y];
  }
}
void solve() {
  int n_nodes, n_queries;
  cin >> n_nodes >> n_queries;
  component_of.resize(n_nodes, -1);
  cp = 0;
  vector<pair<int, pair<int, int>>> ev(n_nodes - 1);
  map<int, long long> ans;
  ans[0] = 0;
  for (int i = 0; i < n_nodes - 1; i++) {
    cin >> ev[i].second.first >> ev[i].second.second >> ev[i].first;
    ev[i].second.first--;
    ev[i].second.second--;
  }
  sort(ev.begin(), ev.end());
  for (int i = 0; i < n_nodes - 1; i++) {
    dsu(ev[i].second.first, ev[i].second.second);
    ans[ev[i].first] = running_ans;
  }
  for (int i = 0; i < n_queries; i++) {
    int q;
    cin >> q;
    auto it = ans.upper_bound(q);
    it--;
    cout << (*it).second << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  while (q-- > 0) {
    solve();
  }
  return 0;
}