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 N = 200005;
int parent[N], sz[N];
long long res[N];
vector<pair<int, int> > edges[N];
void make_set(int v) {
  parent[v] = v;
  sz[v] = 1;
}
int find_set(int v) {
  if (parent[v] == v) return v;
  return parent[v] = find_set(parent[v]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) make_set(i);
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    edges[w].push_back({u, v});
  }
  for (int i = 1; i < N; ++i) {
    res[i] = res[i - 1];
    for (auto p : edges[i]) {
      int a = p.first, b = p.second;
      a = find_set(a);
      b = find_set(b);
      res[i] -= 1ll * sz[a] * (sz[a] - 1) / 2 + 1ll * sz[b] * (sz[b] - 1) / 2;
      if (sz[a] < sz[b]) swap(a, b);
      parent[b] = a;
      sz[a] += sz[b];
      res[i] += 1ll * sz[a] * (sz[a] - 1) / 2;
    }
  }
  while (m--) {
    int q;
    cin >> q;
    cout << res[q] << ' ';
  }
  return 0;
}