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;
using ll = long long;
using pii = pair<int, int>;
const int MN = 2e5 + 5;
int N, Q, par[MN];
ll sz[MN];
struct Edge {
  int u, v, w;
  bool operator<(const Edge &o) const { return w < o.w; }
};
ll cur, ans[MN];
int find(int u) { return u == par[u] ? u : par[u] = find(par[u]); }
void merge(int u, int v) {
  u = find(u), v = find(v);
  if (u == v) return;
  cur += sz[u] * sz[v];
  sz[u] += sz[v];
  par[v] = u;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) par[i] = i, sz[i] = 1;
  vector<Edge> es;
  vector<pii> qs;
  for (int i = 1; i < N; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    es.push_back({u, v, w});
  }
  sort(es.begin(), es.end());
  for (int i = 0; i < Q; i++) {
    int q;
    cin >> q;
    qs.emplace_back(q, i);
  }
  sort(qs.begin(), qs.end());
  int pos = 0;
  for (int i = 0; i < Q; i++) {
    auto [q, idx] = qs[i];
    while (pos < es.size() && es[pos].w <= q) {
      merge(es[pos].u, es[pos].v);
      pos++;
    }
    ans[idx] = cur;
  }
  for (int i = 0; i < Q; i++) cout << ans[i] << " ";
  return 0;
}