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 int MOD = 1e9 + 7;
const int N = 1e6 + 5;
long long int POWER[65];
long long int aonb(long long int a, long long int b) {
  long long int ret = 1;
  while (b) {
    if (b & 1) ret *= a;
    a *= a;
    if (ret >= MOD) ret %= MOD;
    if (a >= MOD) a %= MOD;
    b >>= 1;
  }
  return ret;
}
void precompute() {
  POWER[0] = 1;
  for (int i = 1; i < 63; i++) POWER[i] = POWER[i - 1] << 1LL;
}
long long int res;
vector<int> root, size;
long long int paths(long long int count) {
  return count * 1ll * (count - 1) / 2;
}
int parent(int i) {
  while (i != root[i]) {
    root[i] = root[root[i]];
    i = root[i];
  }
  return i;
}
void perform(int p, int q) {
  p = parent(p);
  q = parent(q);
  if (size[q] > size[p]) {
    swap(p, q);
  }
  root[q] = root[p];
  res -= paths(size[p]);
  res -= paths(size[q]);
  size[p] += size[q];
  res += paths(size[p]);
}
int main() {
  int n, q;
  int a, b, value;
  cin >> n >> q;
  res = 0;
  vector<pair<int, pair<int, int> > > edge(n - 1);
  for (int i = 0; i < n - 1; i++) {
    cin >> edge[i].second.first >> edge[i].second.second >> edge[i].first;
  }
  vector<pair<int, int> > query(q);
  for (int i = 0; i < q; i++) {
    cin >> query[i].first;
    query[i].second = i;
  }
  sort(edge.begin(), edge.end());
  sort(query.begin(), query.end());
  vector<long long int> ans(q);
  root = size = vector<int>(n, 1);
  iota(root.begin(), root.end(), 0);
  int pos = 0;
  for (int i = 0; i < q; i++) {
    while (pos < n - 1 && edge[pos].first <= query[i].first) {
      int p = edge[pos].second.first;
      int q = edge[pos].second.second;
      perform(p, q);
      pos++;
    }
    ans[query[i].second] = res;
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}