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;
int n, m;
vector<pair<int, int> > graph[200100];
pair<int, pair<int, int> > edges[200100];
pair<int, int> arr[200100];
long long size_arr[200100];
int parent[200100];
long long ans[200100];
int find(int x) {
  if (x == parent[x]) {
    return x;
  } else {
    return parent[x] = find(parent[x]);
  }
}
void union_func(int a, int b) {
  int c = find(a);
  int d = find(b);
  if (c != d) {
    parent[d] = c;
    size_arr[c] += size_arr[d];
  }
}
int main() {
  cin >> n >> m;
  int u, v, w;
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> v >> w;
    graph[u].push_back({w, v});
    graph[v].push_back({w, u});
    edges[i - 1].first = w;
    edges[i - 1].second.first = u;
    edges[i - 1].second.second = v;
  }
  for (int i = 0; i < m; i++) {
    cin >> arr[i].first;
    arr[i].second = i;
  }
  sort(arr, arr + m);
  sort(edges, edges + n - 1);
  for (int i = 1; i <= n; i++) {
    parent[i] = i;
    size_arr[i] = 1;
  }
  int j = 0;
  long long ans_now = 0;
  for (int i = 0; i < m; i++) {
    long long max_weight = (long long)arr[i].first;
    int cur_idx = arr[i].second;
    while (j <= n - 2 && (long long)edges[j].first <= max_weight) {
      int edge_1 = edges[j].second.first;
      int edge_2 = edges[j].second.second;
      long long init_size1 =
          size_arr[find(edge_1)] * (size_arr[find(edge_1)] - 1) / 2;
      long long init_size2 =
          size_arr[find(edge_2)] * (size_arr[find(edge_2)] - 1) / 2;
      union_func(edges[j].second.first, edges[j].second.second);
      ans_now += size_arr[find(edge_1)] * (size_arr[find(edge_1)] - 1) / 2 -
                 init_size1 - init_size2;
      j++;
    }
    ans[cur_idx] = ans_now;
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << " ";
  }
}