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>
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long MOD = 998244353;
long long n, m, k, q, l, r, i, j, x, y, ans, a[200001];
long long tt[200001];
struct dsu {
  long long d[200001];
  void make_set(long long h) { d[h] = h; }
  long long get(long long h) {
    if (h == d[h])
      return h;
    else
      return d[h] = get(d[h]);
  }
  void unite_sets(long long a, long long b) {
    a = get(a);
    b = get(b);
    if (a != b) {
      d[b] = a;
      ans -= tt[a] * (tt[a] - 1) / 2;
      ans -= tt[b] * (tt[b] - 1) / 2;
      tt[a] += tt[b];
      ans += (tt[a] * (tt[a] - 1)) / 2;
    }
  }
};
dsu p;
vector<pair<long long, pair<long long, long long> > > v;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    v.push_back(make_pair(z, make_pair(x, y)));
  }
  for (int i = 0; i < m; i++) {
    long long x;
    cin >> x;
    v.push_back(make_pair(x, make_pair(1e18, i)));
  }
  sort(v.begin(), v.end());
  for (int i = 1; i <= n; i++) p.make_set(i), tt[i] = 1;
  for (int i = 0; i < v.size(); i++) {
    if (v[i].second.first == 1e18) {
      a[v[i].second.second] = ans;
    } else {
      p.unite_sets(v[i].second.first, v[i].second.second);
    }
  }
  for (int i = 0; i < m; i++) cout << a[i] << " ";
  return 0;
}