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;
long long parent[200005];
long long size[200005];
void make_set(long long v) {
  parent[v] = v;
  size[v] = 1;
}
long long find_set(long long v) {
  if (v == parent[v]) return v;
  return parent[v] = find_set(parent[v]);
}
void union_sets(long long a, long long b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (size[a] < size[b]) swap(a, b);
    parent[b] = a;
    size[a] += size[b];
  }
}
long long ans[200005];
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long> > > v;
  for (long long i = 1; i <= n; i++) {
    make_set(i);
  }
  for (long long i = 0; i < n - 1; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    v.push_back({z, {x, y}});
  }
  sort((v).begin(), (v).end());
  for (long long i = 0; i < v.size(); i++) {
    pair<long long, pair<long long, long long> > p = v[i];
    long long wei = p.first;
    long long x = size[find_set(p.second.first)];
    long long y = size[find_set(p.second.second)];
    union_sets(p.second.first, p.second.second);
    long long z = size[find_set(p.second.first)];
    ans[wei] += (z * (z - 1)) / 2 - (x * (x - 1)) / 2 - (y * (y - 1)) / 2;
  }
  for (long long i = 1; i <= 200000; i++) {
    ans[i] += ans[i - 1];
  }
  while (m--) {
    long long x;
    cin >> x;
    cout << ans[x] << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  for (long long T = 1; T <= t; T++) {
    solve();
  }
}