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<pair<int, pair<int, int>>> arr;
vector<pair<int, int>> q;
long long n, m, rez = 0;
vector<long long> p, sz;
long long get(long long a) { return 1LL * (a * (a - 1)) / 2; }
int find_set(int v) {
  if (p[v] == v) return v;
  return p[v] = find_set(p[v]);
}
void Union(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (sz[a] < sz[b]) swap(a, b);
    rez -= get(sz[a]);
    rez -= get(sz[b]);
    p[b] = a;
    sz[a] += sz[b];
    rez += get(sz[a]);
  }
}
int main() {
  cin >> n >> m;
  p = sz = vector<long long>(n, 1);
  iota(p.begin(), p.end(), 0);
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    arr.push_back(make_pair(c, make_pair(a, b)));
  }
  sort(arr.begin(), arr.end());
  for (int i = 0; i < m; i++) {
    int a;
    cin >> a;
    q.push_back(make_pair(a, i));
  }
  vector<long long> ans(m, 0);
  sort(q.begin(), q.end());
  int paths = 0;
  for (int i = 0; i < m; i++) {
    while (paths < n - 1 && arr[paths].first <= q[i].first) {
      int u = arr[paths].second.first;
      int v = arr[paths].second.second;
      Union(u, v);
      ++paths;
    }
    ans[q[i].second] = rez;
  }
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}