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;
long long z = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % z;
      b--;
    } else {
      a = (a * a) % z;
      b = b >> 1;
    }
  }
  return res;
}
vector<pair<long long, pair<long long, long long> > > v;
long long par[200001], val[200001] = {0}, siz[200001];
long long find_par(long long x) {
  if (par[x] == x) return x;
  return find_par(par[x]);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long x, y, w;
    for (long long i = 1; i < n; i++) {
      cin >> x >> y >> w;
      v.push_back(make_pair(w, make_pair(x, y)));
    }
    sort(v.begin(), v.end());
    for (long long i = 1; i <= n; i++) {
      par[i] = i;
      siz[i] = 1;
    }
    for (long long i = 0; i < n - 1; i++) {
      x = v[i].second.first;
      y = v[i].second.second;
      w = v[i].first;
      x = find_par(x);
      y = find_par(y);
      if (siz[x] < siz[y]) swap(x, y);
      val[w] += siz[x] * siz[y];
      siz[x] += siz[y];
      par[y] = x;
    }
    for (long long i = 1; i <= 200000; i++) {
      val[i] += val[i - 1];
    }
    while (m--) {
      long long q;
      cin >> q;
      cout << val[q] << " ";
    }
    cout << "\n";
  }
}