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 mod = 1000000007;
const long long INF = 1000010;
long long int res = 0;
std::vector<long long int> parent, rnk;
long long int solve(long long int x) { return x * 1ll * (x - 1) / 2; }
long long int find_set(long long int x) {
  if (x == parent[x]) {
    return x;
  }
  return parent[x] = find_set(parent[x]);
}
void union_sets(long long int a, long long int b) {
  a = find_set(a);
  b = find_set(b);
  if (rnk[a] < rnk[b]) {
    swap(a, b);
  }
  parent[b] = a;
  res -= solve(rnk[a]);
  res -= solve(rnk[b]);
  rnk[a] += rnk[b];
  res += solve(rnk[a]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  parent = vector<long long int>(n);
  rnk = vector<long long int>(n, 1);
  for (int i = 0; i < n; ++i) {
    parent[i] = i;
  }
  std::vector<pair<long long int, pair<long long int, long long int> > > edges(
      n - 1);
  for (int i = 0; i < n - 1; ++i) {
    cin >> edges[i].second.first >> edges[i].second.second >> edges[i].first;
    edges[i].second.first--;
    edges[i].second.second--;
  }
  std::vector<pair<long long int, long long int> > q(m);
  for (int i = 0; i < m; ++i) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  sort(edges.begin(), edges.end());
  std::vector<long long int> ans(m);
  long long int idx = 0;
  for (int i = 0; i < m; ++i) {
    while (idx < n - 1 && edges[idx].first <= q[i].first) {
      long long int u = edges[idx].second.first;
      long long int v = edges[idx].second.second;
      union_sets(u, v);
      idx++;
    }
    ans[q[i].second] = res;
  }
  for (int i = 0; i < m; ++i) {
    cout << ans[i] << " ";
  }
  cout << '\n';
  return 0;
}