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>
const long long N = 500005;
const long long mod = 1e9 + 7;
using namespace std;
long long parent[N];
long long urank[N];
void create(long long x) {
  parent[x] = x;
  urank[x] = 1;
}
long long find(long long x) {
  if (parent[x] != x) {
    parent[x] = find(parent[x]);
  }
  return parent[x];
}
void merge(long long x, long long y) {
  long long xroot = find(x);
  long long yroot = find(y);
  if (urank[xroot] <= urank[yroot]) {
    parent[xroot] = yroot;
    urank[yroot] = urank[yroot] + urank[xroot];
  } else {
    parent[yroot] = xroot;
    urank[xroot] = urank[xroot] + urank[yroot];
  }
}
void solve() {
  long long n, m, k, x = 0, y = 0, c = 0, q, ans = 0;
  cin >> n >> m;
  pair<long long, long long> query[m];
  pair<long long, pair<long long, long long> > a[n - 1];
  for (long long i = 1; i < n + 1; i++) create(i);
  for (long long i = 0; i < n - 1; ++i) {
    cin >> a[i].second.first >> a[i].second.second >> a[i].first;
  }
  for (long long i = 0; i < m; ++i) {
    cin >> query[i].first;
    query[i].second = i;
  }
  sort(a, a + n - 1);
  sort(query, query + m);
  c = 0;
  long long b[m];
  for (int i = 0; i < m; ++i) {
    while (c < n - 1 and a[c].first <= query[i].first) {
      x = find(a[c].second.first);
      y = find(a[c].second.second);
      ans = ans - (urank[x] * (urank[x] - 1)) / 2 -
            (urank[y] * (urank[y] - 1)) / 2;
      merge(x, y);
      x = find(x);
      ans += (urank[x] * (urank[x] - 1)) / 2;
      c++;
    }
    b[query[i].second] = ans;
  }
  for (int i = 0; i < m; ++i) {
    cout << b[i] << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  while (T--) {
    solve();
  }
  return 0;
}