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;
using ll = long long;
using pii = pair<int, int>;
template <typename T, typename S>
ostream& operator<<(ostream& os, pair<T, S> a) {
  os << '(' << a.first << ',' << a.second << ')';
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> v) {
  for (auto x : v) os << x << ' ';
  return os;
}
void debug() { cerr << '\n'; }
template <typename H, typename... T>
void debug(H a, T... b) {
  cerr << a;
  if (sizeof...(b)) cerr << ", ";
  debug(b...);
}
int a[200];
vector<int> g[200];
int n, k;
vector<vector<ll>> dp;
void dfs(int s, int p) {
  dp[s][0] = a[s];
  for (int x : g[s]) {
    if (x == p) continue;
    dfs(x, s);
    vector<ll> ndp(n);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (i + j + 1 > k)
          ndp[min(i, j + 1)] = max(ndp[min(i, j + 1)], dp[s][i] + dp[x][j]);
      }
      if (i + 1 < n) ndp[i + 1] = max(ndp[i + 1], dp[x][i]);
    }
    for (int i = 0; i < n; i++) {
      dp[s][i] = max(dp[s][i], ndp[i]);
    }
  }
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  for (int i = 0; i < (int)(n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dp.resize(n, vector<ll>(n));
  dfs(0, -1);
  cout << *max_element(dp[0].begin(), dp[0].end()) << endl;
}