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 par[200005], sz[200005];
long long root(long long x) {
  return (x == par[x] ? x : (par[x] = root(par[x])));
}
long long ans = 0;
void join(long long x, long long y) {
  ans -= sz[root(y)] * (sz[root(y)] - 1) / 2;
  ans -= sz[root(x)] * (sz[root(x)] - 1) / 2;
  sz[root(x)] += sz[root(y)];
  par[root(y)] = root(x);
  ans += sz[root(x)] * (sz[root(x)] - 1) / 2;
}
long long res[200005];
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long>>> sh;
  for (long long i = 1; i < n; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    sh.push_back({z, {x, y}});
  }
  for (long long i = 1; i <= n; i++) sz[i] = 1, par[i] = i;
  sort(sh.begin(), sh.end());
  long long l = 0;
  vector<pair<long long, long long>> qr;
  for (long long i = 0; i < m; i++) {
    long long x;
    cin >> x;
    qr.push_back({x, i});
  }
  sort(qr.begin(), qr.end());
  for (auto &cv : qr) {
    while (l < sh.size() && sh[l].first <= cv.first) {
      join(sh[l].second.first, sh[l].second.second);
      l++;
    }
    res[cv.second] = ans;
  }
  for (long long i = 0; i < m; i++) cout << res[i] << (i == m - 1 ? '\n' : ' ');
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}