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;
inline void speedUp() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
vector<pair<int, int> > weight[200005];
long long int q[200005], par[200005], cnt[200005];
long long int res;
long long int Find(long long int x) {
  if (par[x] == x) return x;
  return par[x] = Find(par[x]);
}
void Union(long long int a, long long int b) { par[Find(b)] = Find(a); }
void solve() {
  for (int i = 1; i < 200005; i++) {
    for (auto &it : weight[i]) {
      if (Find(it.first) != Find(it.second)) {
        res += cnt[par[it.first]] * cnt[par[it.second]];
        cnt[par[it.first]] += cnt[par[it.second]];
        Union(it.first, it.second);
      }
    }
    q[i] = res;
  }
}
int32_t main() {
  speedUp();
  long long int n, m, u, v, w, x;
  cin >> n >> m;
  for (int i = 1; i < 200005; i++) {
    par[i] = i;
    cnt[i] = 1;
  }
  for (int i = 1; i < n; i++) {
    cin >> u >> v >> w;
    weight[w].push_back({u, v});
  }
  solve();
  for (int i = 1; i <= m; i++) {
    cin >> x;
    cout << q[x] << ' ';
  }
  return 0;
}