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;
int n, m;
int p[2 * 100001], rnk[2 * 100001];
int cnt[2 * 100001];
long long pref[2 * 100001 + 5];
long long ans[2 * 100001 + 5];
vector<pair<int, int> > q;
struct Edge {
  int a, b;
  int w;
  Edge(int a, int b, int w) : a(a), b(b), w(w) {}
};
vector<Edge> e;
bool comp(const Edge &one, const Edge &two) { return one.w < two.w; }
void init() {
  pref[0] = 0;
  for (int i = 1; i <= 2 * 100001; ++i) pref[i] = pref[i - 1] + (i - 1);
}
void dsu_init() {
  for (int i = 0; i < 2 * 100001; ++i) {
    p[i] = i;
    rnk[i] = 0;
    cnt[i] = 1;
  }
}
int dsu_find(int v) {
  if (p[v] == v) return v;
  return p[v] = dsu_find(p[v]);
}
int dsu_meld(int a, int b, int &prev) {
  a = dsu_find(a), b = dsu_find(b);
  if (a == b) return 0;
  if (rnk[a] < rnk[b]) swap(a, b);
  p[b] = a;
  cnt[a] += cnt[b];
  prev = cnt[b];
  if (rnk[a] == rnk[b]) ++rnk[a];
  return cnt[a];
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  init();
  dsu_init();
  cin >> n >> m;
  int a, b, w;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b >> w;
    --a, --b;
    e.emplace_back(a, b, w);
  }
  sort(e.begin(), e.end(), comp);
  for (int i = 0; i < m; ++i) {
    cin >> a;
    q.push_back({a, i});
  }
  sort(q.begin(), q.end());
  long long sum = 0;
  int bnd = 0;
  int prev, now;
  for (int i = 0; i < q.size(); ++i) {
    for (int j = bnd; j < e.size(); ++j) {
      if (e[j].w > q[i].first) break;
      now = dsu_meld(e[j].a, e[j].b, prev);
      sum -= pref[prev];
      sum += (pref[now] - pref[now - prev]);
      ++bnd;
    }
    ans[q[i].second] = sum;
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << " ";
  return 0;
}