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 N = 2e5 + 5;
const int MOD = 1e9 + 7;
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
vector<int> parent;
long long ans1 = 0;
int find(int x) {
  if (parent[x] != x) parent[x] = find(parent[x]);
  return parent[x];
}
void Union(int u, int v, vector<int>& rank) {
  u = find(u);
  v = find(v);
  if (rank[u] < rank[v]) swap(u, v);
  ans1 -= (long long)rank[u] * (rank[u] - 1) / 2;
  ans1 -= (long long)rank[v] * (rank[v] - 1) / 2;
  rank[u] += rank[v];
  ans1 += (long long)(rank[u]) * (rank[u] - 1) / 2;
  parent[v] = u;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<pair<int, pair<int, int>>> adj(n - 1);
    for (int i = 0; i < n - 1; i++) {
      cin >> adj[i].second.first >> adj[i].second.second >> adj[i].first;
      adj[i].second.first--;
      adj[i].second.second--;
    }
    vector<pair<int, int>> q(m);
    for (int i = 0; i < m; i++) {
      cin >> q[i].first;
      q[i].second = i;
    }
    sort(q.begin(), q.end());
    sort(adj.begin(), adj.end());
    int j = 0;
    parent = vector<int>(n, 0);
    vector<int> rank(n, 1);
    for (int i = 0; i < n; i++) parent[i] = i;
    long long ans[m];
    for (int i = 0; i < m; i++) {
      while (j < n - 1 && adj[j].first <= q[i].first) {
        int u = adj[j].second.first;
        int v = adj[j].second.second;
        Union(u, v, rank);
        j++;
      }
      ans[q[i].second] = ans1;
    }
    for (int i = 0; i < m; i++) cout << ans[i] << " ";
    cout << "\n";
  }
}