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 pii = pair<int, int>;
using vi = vector<int>;
using ll = long long;
using vii = vector<pii>;
using vll = vector<ll>;
struct UF {
  vi p, s;
  UF(int n) { p.resize(n), iota(p.begin(), p.end(), 0), s.assign(n, 1); }
  int find(int i) {
    while (p[i] != i) p[i] = p[p[i]], i = p[i];
    return i;
  }
  bool connected(int x, int y) { return find(x) == find(y); }
  bool join(int x, int y) {
    x = find(x), y = find(y);
    if (connected(x, y)) return false;
    if (s[x] < s[y])
      p[x] = y, s[y] += s[x];
    else
      p[y] = x, s[x] += s[y];
    return true;
  }
};
struct edge {
  int u, v, w;
  edge(int _u, int _v, int _w) : u(_u), v(_v), w(_w) {}
  bool operator<(const edge &o) const { return w < o.w; }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<edge> edges;
  for (int _ = int(0); _ < int(n - 1); _++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    edges.emplace_back(u, v, w);
  }
  vii queries(m);
  for (int i = int(0); i < int(m); i++) {
    cin >> queries[i].first;
    queries[i].second = i;
  }
  sort(edges.begin(), edges.end());
  sort(queries.begin(), queries.end());
  vll ans(m);
  UF uf(n);
  int edge_index = 0;
  ll sum = 0;
  for (auto [max_weight, query_index] : queries) {
    while (edge_index < n - 1 && edges[edge_index].w <= max_weight) {
      int u = edges[edge_index].u, v = edges[edge_index].v;
      if (!uf.connected(u, v)) {
        auto pairs = [](ll x) { return x * (x - 1) / 2; };
        sum -= pairs(uf.s[uf.find(u)]);
        sum -= pairs(uf.s[uf.find(v)]);
        uf.join(u, v);
        sum += pairs(uf.s[uf.find(u)]);
      }
      edge_index++;
    }
    ans[query_index] = sum;
  }
  for (int i = int(0); i < int(m); i++) cout << ans[i] << " \n"[i == m - 1];
  return 0;
}