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>
#pragma GCC optimize("Ofast")
using namespace std;
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<long long int>> adj(n);
  for (long long int i = 1; i < n; i++) {
    long long int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<vector<long long int>> dp(n + 1, vector<long long int>(n + 1));
  vector<long long int> dp_temp(n + 1);
  function<void(long long int, long long int)> Dfs = [&](long long int u,
                                                         long long int p) {
    dp[u][0] = a[u];
    for (long long int v : adj[u]) {
      if (v != p) {
        Dfs(v, u);
        dp_temp.assign(n + 1, 0ll);
        for (long long int i = 0; i < n; i++) {
          for (long long int j = max(1ll, k + 1 - i); j < n; j++) {
            dp_temp[min(i, j)] =
                max(dp_temp[min(i, j)], dp[u][i] + dp[v][j - 1]);
          }
        }
        for (long long int i = 1; i < n; i++) {
          dp[u][i] = max(dp[u][i], dp[v][i - 1]);
        }
        for (long long int i = 0; i < n; i++) {
          dp[u][i] = max(dp[u][i], dp_temp[i]);
        }
      }
    }
  };
  Dfs(0, 0);
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    ans = max(ans, dp[0][i]);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  cerr << "Time : "
       << 1000 * ((long double)clock()) / (long double)CLOCKS_PER_SEC << "ms\n";
  return 0;
}