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 MAX = 2e5 + 5;
int parent[MAX];
int depth[MAX];
int get(int a) {
  while (a != parent[a]) a = parent[a];
  return a;
}
void join(int a, int b) {
  a = get(a);
  b = get(b);
  if (a == b) return;
  if (depth[a] > depth[b]) swap(a, b);
  parent[a] = b;
  depth[b] += depth[a];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q, a, b, w;
  cin >> n >> q;
  for (int i = 1; i <= n + 1; i++) {
    parent[i] = i;
    depth[i] = 1;
  }
  vector<pair<int, pair<int, int>>> vp;
  vector<long long> sol(MAX);
  long long cnt = 0;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b >> w;
    vp.push_back({w, {a, b}});
  }
  sort(vp.begin(), vp.end());
  reverse(vp.begin(), vp.end());
  for (int j = 1; j < MAX; j++) {
    while (!vp.empty() && vp.back().first <= j) {
      a = vp.back().second.first;
      b = vp.back().second.second;
      vp.pop_back();
      a = get(a);
      b = get(b);
      if (a == b) continue;
      cnt -= 1ll * depth[a] * (depth[a] - 1) / 2;
      cnt -= 1ll * depth[b] * (depth[b] - 1) / 2;
      join(a, b);
      a = get(a);
      cnt += 1ll * depth[a] * (depth[a] - 1) / 2;
    }
    sol[j] = cnt;
  }
  while (q--) {
    cin >> w;
    cout << sol[w] << " ";
  }
  cout << "\n";
  return 0;
}