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;
template <typename input_type>
input_type get() {
  input_type input;
  cin >> input;
  return input;
}
const int max_sz = (2e2 + 5), inf = 2e9;
int sz, dstnc, wght[max_sz], max_st[max_sz], tmp[max_sz], dp[max_sz][max_sz];
vector<int> nghbr[max_sz];
void dfs(int vrtx, int prnt) {
  max_st[vrtx] = 1;
  dp[vrtx][0] = wght[vrtx];
  for (int ngh : nghbr[vrtx]) {
    if (ngh != prnt) {
      dfs(ngh, vrtx);
      for (int i = 0; i < max(max_st[vrtx], max_st[ngh] + 1); i++) {
        tmp[i] = (-1 * inf);
      }
      for (int i = 0; i < max_st[vrtx]; i++) {
        for (int j = max(0, dstnc - i) + 1; j <= max_st[ngh]; j++) {
          tmp[min(i, j)] = max(tmp[min(i, j)], (dp[vrtx][i] + dp[ngh][j - 1]));
        }
      }
      for (int i = 0; i < max_st[ngh]; i++) {
        tmp[i + 1] = max(tmp[i + 1], dp[ngh][i]);
      }
      for (int i = 0; i < max_st[vrtx]; i++) {
        dp[vrtx][i] = max(dp[vrtx][i], tmp[i]);
      }
      for (int i = max_st[vrtx]; i < max(max_st[vrtx], max_st[ngh] + 1); i++) {
        dp[vrtx][i] = tmp[i];
      }
      max_st[vrtx] = max(max_st[vrtx], max_st[ngh] + 1);
      for (int i = (max_st[vrtx] - 1); i > 0; i--) {
        dp[vrtx][i - 1] = max(dp[vrtx][i - 1], dp[vrtx][i]);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  sz = get<int>();
  dstnc = get<int>();
  for (int i = 0; i < sz; i++) {
    wght[i] = get<int>();
  }
  for (int i = 0; i < (sz - 1); i++) {
    pair<int, int> endpnt = {(get<int>() - 1), (get<int>() - 1)};
    nghbr[endpnt.first].push_back(endpnt.second);
    nghbr[endpnt.second].push_back(endpnt.first);
  }
  dfs(0, -1);
  cout << dp[0][0];
  return 0;
}