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;
#pragma GCC optimization("Ofast")
#pragma GCC optimization("unroll-loops")
#pragma GCC target("avx2,avx,fma")
vector<pair<int, pair<int, int> > > v;
vector<pair<int, int> > q;
int n, m;
vector<int> par;
vector<long long int> siz;
int getp(int a) {
  if (a == par[a]) return a;
  return par[a] = getp(par[a]);
}
long long int res = 0;
void merge(int a, int b) {
  int x = getp(a);
  int y = getp(b);
  if (x != y) {
    if (siz[x] > siz[y]) swap(x, y);
    par[x] = y;
    res -= (siz[x] * (siz[x] - 1)) / 2;
    res -= (siz[y] * (siz[y] - 1)) / 2;
    res += ((siz[x] + siz[y]) * (siz[x] + siz[y] - 1)) / 2;
    siz[y] += siz[x];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int i;
  cin >> n >> m;
  v.resize(n - 1);
  par.resize(n);
  siz.resize(n, 1);
  iota(par.begin(), par.end(), 0);
  q.resize(m);
  for (i = 0; i < n - 1; i++) {
    cin >> v[i].second.first >> v[i].second.second >> v[i].first;
    v[i].second.first--;
    v[i].second.second--;
  }
  for (i = 0; i < m; i++) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  sort(v.begin(), v.end());
  int p = 0;
  vector<long long int> ans(m);
  for (i = 0; i < m; i++) {
    while (p < n - 1 && v[p].first <= q[i].first) {
      merge(v[p].second.first, v[p].second.second);
      p++;
    }
    ans[q[i].second] = res;
  }
  for (i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}