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> prefix(200005, 0);
class DSU {
  vector<long long> parent, rank;

 public:
  DSU(long long n) {
    parent.resize(n);
    rank.resize(n);
    for (int i = 0; i < n; i++) {
      parent[i] = i;
      rank[i] = 1;
    }
  }
  long long get(long long a) {
    if (parent[a] == a) return a;
    return parent[a] = get(parent[a]);
  }
  void union_sets(long long a, long long b) {
    a = get(a);
    b = get(b);
    if (a != b) {
      if (rank[a] < rank[b]) {
        swap(a, b);
      }
      parent[b] = a;
      rank[a] += rank[b];
    }
  }
  long long getRankCount(long long a) {
    a = get(a);
    return rank[a];
  }
};
int main() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> edges;
  for (long long i = 1; i < n; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    edges.push_back({w, u, v});
  }
  sort(edges.begin(), edges.end());
  DSU g(n + 1);
  for (long long i = 0; i < n - 1; i++) {
    long long u = edges[i][1];
    long long v = edges[i][2];
    long long w = edges[i][0];
    prefix[w] += g.getRankCount(u) * g.getRankCount(v);
    g.union_sets(u, v);
  }
  for (long long i = 1; i < 200005; i++) {
    prefix[i] += prefix[i - 1];
  }
  while (m--) {
    long long w;
    cin >> w;
    cout << prefix[w] << " ";
  }
  cout << endl;
  return 0;
}