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;
const int M = 200010;
int par[M];
int sz[M];
struct edge {
  int u, v, w;
  bool operator<(const edge &e) const { return w < e.w; }
};
edge E[M];
pair<int, int> q[M];
long long out[M];
int f(int u) { return par[u] = par[u] == u ? u : f(par[u]); }
int add(int u, int v) {
  u = f(u);
  v = f(v);
  if (u == v) return 0;
  par[v] = u;
  sz[u] += sz[v];
  return sz[u];
}
long long get(int n) {
  if (n <= 0) return 0;
  return ((long long)n * (long long)(n - 1)) / 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    cin >> E[i].u >> E[i].v >> E[i].w;
  }
  for (int i = 0; i < m; i++) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(E, E + n - 1);
  sort(q, q + m);
  for (int i = 1; i <= n; i++) par[i] = i, sz[i] = 1;
  long long ans = 0;
  int j = 0;
  for (int i = 0; i < m; i++) {
    while (j < n - 1) {
      if (E[j].w > q[i].first) break;
      int u = f(E[j].u), v = f(E[j].v);
      ans -= (get(sz[u]) + get(sz[v]));
      ans += get(add(u, v));
      j++;
    }
    out[q[i].second] = ans;
  }
  for (int i = 0; i < m; i++) cout << out[i] << " ";
  cout << endl;
  return 0;
}