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;
vector<long long int> P(200005);
vector<long long int> R(200005);
long long int get(long long int x) {
  if (P[x] != x) return P[x] = get(P[x]);
  return P[x];
}
long long int res;
long long int solve(long long int x) { return (x * (x - 1)) / 2; }
void merge(long long int x, long long int y) {
  long long int u = get(x);
  long long int v = get(y);
  if (R[u] < R[v]) swap(u, v);
  res -= solve(R[u]);
  res -= solve(R[v]);
  R[u] += R[v];
  res += solve(R[u]);
  P[v] = u;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  res = 0;
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) P[i] = i, R[i] = 1;
  vector<pair<long long int, pair<long long int, long long int>>> E(n - 1);
  for (long long int i = 0; i < n - 1; i++) {
    long long int u, v, w;
    cin >> u >> v >> w;
    E[i] = {w, {--u, --v}};
  }
  vector<pair<long long int, long long int>> Q(m);
  for (long long int i = 0; i < m; i++) cin >> Q[i].first, Q[i].second = i;
  sort(E.begin(), E.end());
  sort(Q.begin(), Q.end());
  long long int pos = 0;
  vector<long long int> Sol(m);
  for (long long int i = 0; i < m; i++) {
    while (pos < n - 1 && E[pos].first <= Q[i].first) {
      long long int u = E[pos].second.first;
      long long int v = E[pos].second.second;
      merge(u, v);
      pos++;
    }
    Sol[Q[i].second] = res;
  }
  for (long long int i = 0; i < m; i++) cout << Sol[i] << " ";
  cout << endl;
}