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 int pow_mod(long long int a, long long int b) {
  long long int res = 1;
  while (b != 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res;
}
long long int sz[200005];
long long int par[200005];
void initialize(long long int n) {
  for (long long int i = 0; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
}
long long int find(long long int x) {
  if (x == par[x]) return x;
  return par[x] = find(par[x]);
}
void merge(long long int x, long long int y) {
  if (sz[x] < sz[y]) {
    swap(x, y);
  }
  sz[x] += sz[y];
  par[y] = x;
}
long long int nc2(long long int x) { return (x * (x - 1)) / 2; }
void solve() {
  long long int n, m;
  cin >> n >> m;
  vector<pair<long long int, pair<long long int, long long int> > > v;
  long long int x[n], y[n], w[n];
  for (long long int i = 0; i < n - 1; i++) {
    cin >> x[i] >> y[i] >> w[i];
    v.push_back(make_pair(w[i], make_pair(0LL, i)));
  }
  long long int ans[m];
  for (long long int i = 0; i < m; i++) {
    long long int xx;
    cin >> xx;
    v.push_back(make_pair(xx, make_pair(1LL, i)));
  }
  if (!v.empty()) sort(v.begin(), v.end());
  long long int res = 0;
  initialize(n);
  for (long long int i = 0; i < v.size(); i++) {
    long long int type = v[i].second.first;
    if (type == 0) {
      long long int xx, yy;
      xx = x[v[i].second.second];
      yy = y[v[i].second.second];
      xx = find(xx);
      yy = find(yy);
      res -= nc2(sz[xx]);
      res -= nc2(sz[yy]);
      res += nc2(sz[xx] + sz[yy]);
      merge(xx, yy);
    } else {
      ans[v[i].second.second] = res;
    }
  }
  for (long long int i = 0; i < m; i++) cout << ans[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) solve();
  return 0;
}