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 long long mod = (1 ? 1e9 + 7 : 998244353);
const int N = 2e3 + 2;
const int M = 1 << 17;
const int inf = 2e9;
const long long linf = 1e18;
vector<int> adj[N];
int dp[N][N];
int n, k;
void dfs(int s, int e) {
  vector<int> c;
  for (auto u : adj[s]) {
    if (u == e) continue;
    dfs(u, s);
    dp[s][0] += dp[u][k - 1];
    c.push_back(u);
  }
  for (int d = 1; d < n; d++) {
    int ans;
    for (int ii = 0; ii < c.size(); ii++) {
      int i = c[ii];
      ans = dp[i][d - 1];
      for (int jj = 0; jj < c.size(); jj++) {
        int j = c[jj];
        if (i == j) continue;
        ans += dp[j][max(d - 1, k - d - 1)];
      }
      dp[s][d] = max(dp[s][d], ans);
    }
  }
  for (int d = n - 2; ~d; d--) dp[s][d] = max(dp[s][d], dp[s][d + 1]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n >> k;
  k++;
  for (int i = 1; i <= n; i++) cin >> dp[i][0];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  cout << dp[1][0] << '\n';
  return 0;
}