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 maxn = 201;
void dfs(int idx, int par, vector<vector<int>>& adj, vector<vector<int>>& dp,
         vector<int>& vec, int k) {
  dp[idx][0] = vec[idx];
  for (int v : adj[idx]) {
    if (v != par) {
      dfs(v, idx, adj, dp, vec, k);
    }
  }
  for (int d = 0; d <= k; d++) {
    if (d == 0) {
      for (int to : adj[idx]) {
        if (to != par) {
          dp[idx][d] += dp[to][k - 1];
        }
      }
    } else {
      int need = max(d, k - d) - 1;
      for (int to : adj[idx]) {
        if (to != par) {
          int current = dp[to][d - 1];
          for (int other : adj[idx]) {
            if (other != to && other != par) {
              current += dp[other][need];
            }
          }
          dp[idx][d] = max(dp[idx][d], current);
        }
      }
    }
  }
  for (int i = k - 1; i >= 0; i--) {
    dp[idx][i] = max(dp[idx][i], dp[idx][i + 1]);
  }
}
int32_t main() {
  int n, k;
  cin >> n >> k;
  k++;
  vector<int> vec(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> vec[i];
  }
  vector<vector<int>> adj(n + 1);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  vector<vector<int>> dp(maxn, vector<int>(maxn, 0));
  dfs(1, 0, adj, dp, vec, k);
  cout << dp[1][0];
  return 0;
}