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;
struct UF {
  vector<long long> e;
  UF(long long n) : e(n, -1) {}
  bool same_set(long long a, long long b) { return find(a) == find(b); }
  long long size(long long x) { return -e[find(x)]; }
  long long find(long long x) { return e[x] < 0 ? x : e[x] = find(e[x]); }
  bool join(long long a, long long b) {
    a = find(a), b = find(b);
    if (a == b) return false;
    if (e[a] > e[b]) swap(a, b);
    e[a] += e[b];
    e[b] = a;
    return true;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long>>> edges(n - 1);
  for (long long i = (0); i < (n - 1); ++i) {
    cin >> edges[i].second.first >> edges[i].second.second >> edges[i].first;
  }
  sort(edges.begin(), edges.end());
  vector<pair<long long, long long>> queries(m);
  for (long long i = (0); i < (m); ++i) {
    cin >> queries[i].first;
    queries[i].second = i;
  }
  sort(queries.begin(), queries.end());
  UF uf(n + 1);
  long long j = 0;
  long long ans = 0;
  vector<long long> out(m);
  for (long long i = (0); i < (m); ++i) {
    long long w = queries[i].first;
    while (j < n - 1 && edges[j].first <= w) {
      ++j;
      if (!uf.same_set(edges[j - 1].second.first, edges[j - 1].second.second)) {
        ans += uf.size(edges[j - 1].second.first) *
               uf.size(edges[j - 1].second.second);
        uf.join(edges[j - 1].second.first, edges[j - 1].second.second);
      }
    }
    out[queries[i].second] = ans;
  }
  for (auto& a : out) cout << a << " ";
  cout << endl;
}