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, k;
int a[210];
int siz[210];
int dp[210][210];
vector<int> neighbors[210];
void dfs(int root, int fa) {
  int sons = 0;
  vector<int> sv;
  siz[root] = 0;
  for (int v : neighbors[root]) {
    if (v != fa) {
      sv.push_back(v);
      sons++;
      dfs(v, root);
      siz[root] = max(siz[root], siz[v] + 1);
    }
  }
  if (sons == 0) {
    siz[root] = 0;
    dp[root][0] = a[root];
    return;
  }
  dp[root][0] = a[root];
  for (int v : sv) {
    dp[root][0] += dp[v][k];
  }
  for (int dep = siz[root]; dep >= 1; --dep) {
    for (int v : sv) {
      int Max = 0;
      Max += dp[v][dep - 1];
      for (int other : sv) {
        if (other == v) continue;
        Max += (k - dep >= 0) ? dp[other][max(dep - 1, k - dep)] : 0;
      }
      dp[root][dep] = max(dp[root][dep], Max);
      dp[root][dep - 1] = max(dp[root][dep - 1], dp[root][dep]);
    }
  }
  return;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    neighbors[u].push_back(v);
    neighbors[v].push_back(u);
  }
  memset(dp, 0, sizeof(dp));
  dfs(1, 0);
  int ans = 0;
  cout << dp[1][0] << endl;
}