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 ed {
  int u, v, w;
  ed(int u, int v, int w) : u(u), v(v), w(w) {}
};
struct que {
  int q, id;
  long long val;
  que(int q, int id, long long val) : q(q), id(id), val(val){};
};
int parent[((int)2e5 + 10)], siz[((int)2e5 + 10)];
int find_set(int v) {
  if (v == parent[v]) return v;
  return parent[v] = find_set(parent[v]);
}
void make_set(int v) {
  parent[v] = v;
  siz[v] = 1;
}
void union_sets(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (siz[a] < siz[b]) swap(a, b);
    parent[b] = a;
    siz[a] += siz[b];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<ed> edges;
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    edges.push_back(ed(u, v, w));
  }
  sort(edges.begin(), edges.end(), [](ed a, ed b) { return a.w < b.w; });
  vector<que> queries;
  for (int i = 0; i < m; i++) {
    int q;
    cin >> q;
    queries.push_back(que(q, i, 0));
  }
  sort(queries.begin(), queries.end(), [](que a, que b) { return a.q < b.q; });
  auto it = edges.begin();
  for (int i = 0; i < n; i++) make_set(i + 1);
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    while (it != edges.end() && it->w <= queries[i].q) {
      ans += ((long long)siz[find_set(it->u)] * siz[find_set(it->v)]);
      union_sets(it->u, it->v);
      it++;
    }
    queries[i].val = ans;
  }
  sort(queries.begin(), queries.end(),
       [](que a, que b) { return a.id < b.id; });
  for (auto q : queries) cout << q.val << " ";
}