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;
const int N = int(2.1e5);
vector<pair<int, int>> g[N];
int n, q;
ll res = 0;
class dsu {
 public:
  using T = int;
  vector<int> p;
  vector<ll> sz;
  int n;
  dsu(int _n) : n(_n) {
    p.resize(n);
    sz.resize(n);
    iota(p.begin(), p.end(), 0);
    fill(sz.begin(), sz.end(), 1);
  }
  inline T get(T x) { return (x == p[x] ? x : (p[x] = get(p[x]))); }
  inline bool unite(T x, T y) {
    T px = get(x);
    T py = get(y);
    if (px != py) {
      if (sz[px] < sz[py]) {
        swap(px, py);
      }
      p[py] = px;
      res -= (sz[px] * (sz[px] - 1)) / 2;
      res -= (sz[py] * (sz[py] - 1)) / 2;
      sz[px] += sz[py];
      res += (sz[px] * (sz[px] - 1)) / 2;
      return true;
    }
    return false;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> q;
  vector<pair<int, pair<int, int>>> ed(n - 1);
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    ed[i] = {w, {u, v}};
  }
  sort(ed.begin(), ed.end());
  vector<pair<int, int>> qr(q);
  for (int i = 0; i < q; ++i) {
    int x;
    cin >> x;
    qr[i] = {x, i};
  }
  sort(qr.begin(), qr.end());
  vector<ll> ans(q);
  int j = 0;
  dsu d(n + 1);
  for (int i = 0; i < q; ++i) {
    while (j < n - 1 && ed[j].first <= qr[i].first) {
      d.unite(ed[j].second.first, ed[j].second.second);
      j++;
    }
    ans[qr[i].second] = res;
  }
  for (ll x : ans) {
    cout << x << ' ';
  }
  cout << '\n';
  return 0;
}