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;
template <typename T>
void _dbg(char const *s, const T &t) {
  cout << s << "=" << t << endl;
}
template <typename T, typename... TA>
void _dbg(char const *s, const T &t, const TA &...ta) {
  while (*s != ',') cout << *s++;
  cout << "=" << t << ",";
  _dbg(s + 1, ta...);
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream &print_iterable(ostream &os, T beg, T end) {
  os << '[';
  for (auto it = beg; it != end; ++it) {
    if (it != beg) os << ", ";
    os << *it;
  }
  return os << ']';
}
template <typename T, size_t N>
ostream &operator<<(ostream &os, const array<T, N> &x) {
  return print_iterable(os, x.begin(), x.end());
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &x) {
  return print_iterable(os, x.begin(), x.end());
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &x) {
  return print_iterable(os, x.begin(), x.end());
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &x) {
  return print_iterable(os, x.begin(), x.end());
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &x) {
  return print_iterable(os, x.begin(), x.end());
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const map<T1, T2> &x) {
  return print_iterable(os, x.begin(), x.end());
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const unordered_map<T1, T2> &x) {
  return print_iterable(os, x.begin(), x.end());
}
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<ld, ld>;
const int N = 202;
int ar[N];
vector<int> gr[N];
int dp[N][N * 2], n, k;
void Dfs(int cur, int par = 0) {
  vector<int> sum(N + 1);
  for (auto &x : gr[cur]) {
    if (x == par) {
      continue;
    }
    Dfs(x, cur);
    for (int i = 0; i < N; ++i) {
      sum[i] += dp[x][i];
    }
  }
  dp[cur][0] = ar[cur] + sum[k];
  for (int i = 1; i < N; ++i) {
    int other = max(i - 1, k - i);
    for (auto &x : gr[cur]) {
      if (x == par) {
        continue;
      }
      dp[cur][i] = max(dp[cur][i], dp[x][i - 1] + sum[other] - dp[x][other]);
    }
  }
  for (int i = N - 1; i >= 0; --i) {
    dp[cur][i] = max(dp[cur][i], dp[cur][i + 1]);
  }
}
void Solve() {
  int a, b;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> ar[i];
  }
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  Dfs(1);
  cout << dp[1][0] << endl;
}
int main() {
  Solve();
  return 0;
}