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 edge {
  int u, v, w;
  bool operator<(const edge x) const { return w < x.w; }
  edge(int u, int v, int w) : u(u), v(v), w(w){};
};
vector<edge> e;
int fa[200001];
vector<int> qb(200001), qa(200001), siz(200001);
long long sum[200001];
int find(int x) {
  if (fa[x] == x)
    return x;
  else
    return fa[x] = find(fa[x]);
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  int u, v, w;
  for (int i = 1; i < n; i++) {
    cin >> u >> v >> w;
    e.push_back(edge(u, v, w));
  }
  int mw;
  for (int i = 0; i < m; i++) {
    cin >> mw;
    qb[i] = qa[i] = mw;
  }
  sort(qa.begin(), qa.begin() + m);
  sort(e.begin(), e.end());
  for (int i = 1; i <= n; i++) {
    fa[i] = i;
    siz[i] = 1;
  }
  int k = 0;
  long long add = 0;
  for (int i = 0; i < m; i++) {
    while (k <= n - 2 && e[k].w <= qa[i]) {
      int f1 = find(e[k].u), f2 = find(e[k].v);
      add += 1LL * siz[f1] * siz[f2];
      fa[f2] = f1;
      siz[f1] += siz[f2];
      k++;
    }
    sum[qa[i]] = add;
  }
  for (int i = 0; i < m - 1; i++) {
    cout << sum[qb[i]] << " ";
  }
  cout << sum[qb[m - 1]];
  return 0;
}