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 t {
  long long u;
  long long v;
  long long w;
};
bool uporedi(t a, t b) { return (a.w < b.w); }
template <long long size>
struct dsu {
  long long p[size + 10];
  long long s[size + 10];
  void init() {
    for (long long i = 0; i < size + 10; ++i) {
      p[i] = i;
      s[i] = 1;
    }
  }
  long long find(long long x) {
    if (x == p[x]) return x;
    return p[x] = find(p[x]);
  }
  void make(long long x) { p[x] = x; }
  void spoj(long long a, long long b) {
    a = find(a);
    b = find(b);
    if (a != b) {
      if (s[a] < s[b]) swap(a, b);
      p[b] = a;
      s[a] += s[b];
    }
  }
};
dsu<131072 * 4> d;
long long f(long long x) { return (x * (x - 1)) / 2; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  d.init();
  long long n, m;
  cin >> n >> m;
  vector<t> e(n - 1);
  for (long long i = 0; i < n - 1; ++i) {
    long long u, v, w;
    cin >> u >> v >> w;
    e[i].u = u;
    e[i].v = v;
    e[i].w = w;
  }
  vector<long long> q(m);
  for (auto& i : q) cin >> i;
  vector<long long> tmp = q;
  sort(e.begin(), e.end(), uporedi);
  sort(q.begin(), q.end());
  long long cur = 0, sol = 0;
  vector<long long> res(m);
  for (auto& g : e) {
    while (cur < m && g.w > q[cur]) {
      ++cur;
      res[cur] = res[cur - 1];
    }
    if (cur == m) break;
    long long p1 = d.find(g.u), p2 = d.find(g.v);
    long long s1 = d.s[p1], s2 = d.s[p2];
    sol -= f(s1);
    sol -= f(s2);
    sol += f(s1 + s2);
    res[cur] = sol;
    d.spoj(g.u, g.v);
  }
  for (long long i = cur + 1; i < m; ++i) res[i] = res[cur];
  map<long long, long long> mapa;
  for (long long i = 0; i < m; ++i) {
    mapa[q[i]] = res[i];
  }
  for (auto& i : tmp) {
    cout << mapa[i] << ' ';
  }
  cout << '\n';
}