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[200110];
long long s[200110];
long long form(long long a) { return (a * (a - 1)) / 2; }
long long parent(long long a) {
  if (par[a] == a) {
    return a;
  }
  par[a] = parent(par[a]);
  return par[a];
}
void uni(long long a, long long b, long long& ans) {
  a = parent(a);
  b = parent(b);
  if (a == b) return;
  if (s[a] < s[b]) {
    swap(a, b);
  }
  ans -= (form(s[a]) + form(s[b]));
  s[a] += s[b];
  ans += form(s[a]);
  par[b] = a;
}
bool custom(pair<pair<long long, long long>, long long> p1,
            pair<pair<long long, long long>, long long> p2) {
  return p1.second < p2.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  ;
  long long n, m;
  cin >> n >> m;
  pair<pair<long long, long long>, long long> p[n];
  for (long long i = 0; i < n - 1; i++) {
    cin >> p[i].first.first >> p[i].first.second >> p[i].second;
  }
  sort(p, p + (n - 1), custom);
  for (long long i = 1; i <= n + 10; i++) {
    par[i] = i;
    s[i] = 1;
  }
  long long q[m];
  long long val = 0;
  for (long long i = 0; i < m; i++) {
    cin >> q[i];
    val = max(val, q[i]);
  }
  long long g[200110] = {};
  long long ans = 0;
  for (long long i = 0; i < n - 1; i++) {
    uni(p[i].first.first, p[i].first.second, ans);
    g[p[i].second] = ans;
  }
  long long r = g[0];
  for (long long i = 0; i <= val + 1; i++) {
    if (g[i] == 0) {
      g[i] = r;
    } else {
      r = g[i];
    }
  }
  for (long long i = 0; i < m; i++) {
    cout << g[q[i]] << " ";
  }
}