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;
mt19937 rng(1477);
const int MAXN = 2e5 + 10;
const ll INF = 1e9 + 10;
ll res;
vector<int> dsu_size, dsu_parent;
void dsu_build(int n) {
  dsu_size.resize(n, 1);
  dsu_parent.resize(n);
  iota(dsu_parent.begin(), dsu_parent.end(), 0);
}
int dsu_get_parent(int a) {
  if (a == dsu_parent[a]) return a;
  return dsu_parent[a] = dsu_get_parent(dsu_parent[a]);
}
void dsu_unite(int a, int b) {
  a = dsu_get_parent(a);
  b = dsu_get_parent(b);
  if (a != b) {
    ll sza = dsu_size[a];
    ll szb = dsu_size[b];
    res += sza * szb;
    if (sza < szb) swap(a, b);
    dsu_size[a] += dsu_size[b];
    dsu_parent[b] = a;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<tuple<int, int, int>> edges(n - 1);
  for (auto &e : edges) cin >> get<1>(e) >> get<2>(e) >> get<0>(e);
  sort(edges.begin(), edges.end());
  vector<pair<int, int>> queries(m);
  for (int i = 0; i < m; ++i) {
    cin >> queries[i].first;
    queries[i].second = i;
  }
  sort(queries.begin(), queries.end());
  vector<ll> ans(m);
  dsu_build(n);
  for (int q = 0, i = 0; q < m; ++q) {
    while (i < edges.size() && get<0>(edges[i]) <= queries[q].first) {
      dsu_unite(get<1>(edges[i]) - 1, get<2>(edges[i]) - 1);
      ++i;
    }
    ans[queries[q].second] = res;
  }
  for (auto &e : ans) cout << e << " ";
}