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 link[2 * 100000 + 1], sizet[2 * 100000 + 1], n, m, res = 0;
long long find(long long x) {
  while (x != link[x]) x = link[x];
  return x;
}
bool same(long long i, long long j) { return find(i) == find(j); }
void unite(long long i, long long j) {
  i = find(i);
  j = find(j);
  if (sizet[i] < sizet[j]) swap(i, j);
  if (!same(i, j)) {
    res -= sizet[i] * (sizet[i] - 1) / 2;
    res -= sizet[j] * (sizet[j] - 1) / 2;
    sizet[i] += sizet[j];
    res += sizet[i] * (sizet[i] - 1) / 2;
    link[j] = i;
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long> > > A(n);
  A[n - 1].first = 232213;
  vector<pair<long long, long long> > q(m);
  vector<long long> ans(n + 3);
  for (long long i = 1; i <= n - 1; i++) {
    cin >> A[i - 1].second.first >> A[i - 1].second.second >> A[i - 1].first;
    link[i] = i;
    sizet[i] = 1;
  }
  link[n] = n;
  sizet[n] = 1;
  for (long long j = 0; j <= m - 1; j++) {
    cin >> q[j].first;
    q[j].second = j + 1;
  }
  sort(A.begin(), A.end());
  sort(q.begin(), q.end());
  long long cur = 0;
  for (long long i = 0; i <= m - 1; i++) {
    long long l = q[i].first;
    while (A[cur].first <= l && cur <= n - 1) {
      if (!same(A[cur].second.first, A[cur].second.second)) {
        unite(A[cur].second.first, A[cur].second.second);
      }
      cur++;
    }
    ans[q[i].second] = res;
  }
  for (long long i = 1; i <= m; i++) {
    cout << ans[i] << " ";
  }
}