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 node {
  long long parent;
  long long size;
};
node arr[200005];
void initialise() {
  for (long long i = 0; i < 200005; i++) {
    arr[i].parent = i;
    arr[i].size = 1;
  }
}
long long find_parent(long long i) {
  if (arr[i].parent != i) {
    arr[i].parent = find_parent(arr[i].parent);
  }
  return arr[i].parent;
}
void Union(long long x, long long y) {
  long long x_par = find_parent(x);
  long long y_par = find_parent(y);
  if (arr[x_par].size < arr[y_par].size) {
    arr[x_par].parent = y_par;
    arr[y_par].size += arr[x_par].size;
  } else {
    arr[y_par].parent = x_par;
    arr[x_par].size += arr[y_par].size;
  }
}
long long addEdge(long long a, long long b) {
  long long x = find_parent(a);
  long long y = find_parent(b);
  long long val = arr[x].size * arr[y].size;
  Union(x, y);
  return val;
}
long long ans[200005] = {0};
vector<pair<long long, pair<long long, long long> > > vect;
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  initialise();
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    vect.push_back({w, {u, v}});
  }
  sort(vect.begin(), vect.end());
  for (long long i = 0; i < vect.size(); i++) {
    pair<long long, long long> p = vect[i].second;
    long long wt = vect[i].first;
    ans[wt] += addEdge(p.first, p.second);
  }
  for (long long i = 1; i < 200005; i++) {
    ans[i] += ans[i - 1];
  }
  for (long long i = 0; i < m; i++) {
    long long a;
    cin >> a;
    cout << ans[a] << endl;
  }
  return 0;
}