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 = 3e5 + 100;
const long long MOD = 1e9 + 7;
vector<pair<long long, pair<long long, long long>>> ed;
vector<pair<long long, long long>> quer;
long long par[N];
long long sz[N];
long long ans;
long long res[N];
void make_set(long long v) {
  par[v] = v;
  sz[v] = 1;
}
long long find(long long v) {
  if (par[v] == v)
    return v;
  else
    return par[v] = find(par[v]);
}
void merge(long long u, long long v) {
  u = find(u);
  v = find(v);
  if (u != v) {
    if (sz[u] < sz[v]) {
      swap(u, v);
    }
    par[v] = u;
    ans += (sz[u] * sz[v]);
    sz[u] += sz[v];
  }
}
int main() {
  long long n, m;
  cin >> n >> m;
  for (int i = 2; i <= n; i++) {
    long long u, v, d;
    cin >> u >> v >> d;
    ed.push_back(make_pair(d, make_pair(u, v)));
  }
  for (int i = 1; i <= n; i++) make_set(i);
  for (int i = 1; i <= m; i++) {
    long long x;
    cin >> x;
    quer.push_back(make_pair(x, i));
  }
  sort(ed.begin(), ed.end());
  sort(quer.begin(), quer.end());
  int pos = 0;
  for (int i = 0; i < quer.size(); i++) {
    while (pos < ed.size() && ed[pos].first <= quer[i].first) {
      long long u = ed[pos].second.first;
      long long v = ed[pos].second.second;
      merge(u, v);
      pos++;
    }
    res[quer[i].second] = ans;
  }
  for (int i = 1; i <= m; i++) cout << res[i] << " ";
  cout << endl;
  return 0;
}