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;
const long long N = 2e5 + 5;
long long par[N];
long long sz[N];
long long n, m;
long long find(long long x) {
  if (x == par[x]) return x;
  return par[x] = find(par[x]);
}
void join(long long x, long long y) {
  long long a = find(x);
  long long b = find(y);
  if (sz[b] > sz[a]) swap(a, b);
  par[b] = a;
  sz[a] += sz[b];
  return;
}
long long nC2(long long x) { return (x * (x - 1)) / 2; }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < N; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long> > > ed(n - 1);
  for (long long i = 0; i < n - 1; i++)
    cin >> ed[i].second.first >> ed[i].second.second >> ed[i].first;
  sort(ed.begin(), ed.end());
  vector<pair<long long, long long> > query(m);
  for (long long i = 0; i < m; i++) cin >> query[i].first, query[i].second = i;
  sort(query.begin(), query.end());
  vector<long long> ans(m);
  long long j = 0;
  long long s = 0;
  for (long long i = 0; i < m; i++) {
    long long x = query[i].first;
    while (j < n - 1 && ed[j].first <= x) {
      long long pu = find(ed[j].second.first);
      long long pv = find(ed[j].second.second);
      s -= (nC2(sz[pu]) + nC2(sz[pv]));
      join(pu, pv);
      pu = find(ed[j].second.first);
      s += nC2(sz[pu]);
      j++;
    }
    ans[query[i].second] = s;
  }
  for (long long i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}