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("O3", "unroll-all-loops")
#pragma GCC target("avx2")
using namespace std;
template <class T, class U>
inline void checkmin(T &x, U y) {
  if (y < x) x = y;
}
template <class T, class U>
inline void checkmax(T &x, U y) {
  if (y > x) x = y;
}
template <class T, class U>
inline bool ifmax(T &x, U y) {
  if (y > x) return x = y, true;
  return false;
}
template <class T, class U>
inline bool ifmin(T &x, U y) {
  if (y < x) return x = y, true;
  return false;
}
template <class T>
inline void sort(T &a) {
  sort(a.begin(), a.end());
}
template <class T>
inline void rsort(T &a) {
  sort(a.rbegin(), a.rend());
}
template <class T>
inline void reverse(T &a) {
  reverse(a.begin(), a.end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
  return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) str >> i;
  return str;
}
template <class T>
inline T sorted(T a) {
  sort(a);
  return a;
}
int n, k;
int a[200];
vector<pair<int, int> > G;
int dp[201][200];
int sumdp[201];
void countdp(int v, int p = -1) {
  vector<int> sons;
  for (auto i : G) {
    if (i.first == v && i.second != p) {
      countdp(i.second, v);
      sons.push_back(i.second);
    } else if (i.second == v && i.first != p) {
      countdp(i.first, v);
      sons.push_back(i.first);
    }
  }
  dp[0][v] = a[v];
  for (auto i : sons) dp[0][v] += dp[k][i];
  for (int h = 0; h < 100; ++h) {
    int other = max(h, max(k / 2, k - h - 1));
    int sum = 0;
    for (auto i : sons) sum += dp[other][i];
    for (auto i : sons) checkmax(dp[h + 1][v], sum - dp[other][i] + dp[h][i]);
  }
  for (int i = n - 1; i >= 0; --i) checkmax(dp[i][v], dp[i + 1][v]);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  srand(94385);
  cin >> n >> k;
  checkmin(k, n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    G.emplace_back(u, v);
  }
  countdp(0);
  int ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= n; ++j) checkmax(ans, dp[j][i]);
  cout << ans;
  return 0;
}