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<int> f;
vector<int> cnt;
vector<tuple<int, int, int> > tr;
vector<tuple<int, int> > cw;
vector<long long> answ;
int n, m;
int getF(int u) { return f[u] == u ? u : f[u] = getF(f[u]); }
void read_data() {
  cin >> n >> m;
  cnt.resize(n + 1);
  f.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    f[i] = i;
    cnt[i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    tr.emplace_back(w, u, v);
  }
  cw.reserve(m);
  for (int i = 0; i < m; i++) {
    int w;
    scanf("%d", &w);
    cw.emplace_back(w, i);
  }
  sort(tr.begin(), tr.end());
  sort(cw.begin(), cw.end());
}
void deal() {
  int i = 0;
  long long ans = 0;
  answ.resize(m);
  for (int k = 0; k < m; k++) {
    while (i < n - 1 && get<0>(tr[i]) <= get<0>(cw[k])) {
      int fu = getF(get<1>(tr[i]));
      int fv = getF(get<2>(tr[i]));
      int cfu = cnt[fu];
      int cfv = cnt[fv];
      ans += 1LL * (cfu + cfv) * (cfu + cfv - 1) / 2 -
             1LL * cfu * (cfu - 1) / 2 - 1LL * cfv * (cfv - 1) / 2;
      f[fu] = fv;
      cnt[fv] += cnt[fu];
      i++;
    }
    answ[get<1>(cw[k])] = ans;
  }
  for (int i = 0; i < m; i++) printf("%I64d ", answ[i]);
  cout << endl;
}
int main() {
  read_data();
  deal();
  return 0;
}