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;
const int maxn = 2e5 + 69;
const long long mod = 1e9 + 7;
const long long lim = (1LL << 31) - 1;
long long par[maxn], S[maxn], kq[maxn];
array<int, 2> a[maxn];
int Find_par(int u) {
  if (par[u] != u) par[u] = Find_par(par[u]);
  return par[u];
}
void Link(int u, int v) {
  if (S[u] < S[v]) swap(u, v);
  S[u] += S[v];
  par[v] = u;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<array<int, 3>> edges;
  for (int i = 1; i < n; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    edges.push_back({w, u, v});
  }
  for (int i = 1; i <= n; i++) {
    S[i] = 1;
    par[i] = i;
  }
  for (int i = 1; i <= m; i++) {
    cin >> a[i][0];
    a[i][1] = i;
  }
  sort(a + 1, a + m + 1);
  sort(edges.begin(), edges.end());
  int dem = 1;
  long long ans = 0;
  for (auto j : edges) {
    while (j[0] > a[dem][0] && dem <= m) {
      kq[a[dem][1]] = ans;
      dem++;
    }
    int paru = Find_par(j[1]);
    int parv = Find_par(j[2]);
    if (paru == parv) continue;
    ans = ans - S[paru] * (S[paru] - 1) / 2 - S[parv] * (S[parv] - 1) / 2 +
          (S[parv] + S[paru]) * (S[parv] + S[paru] - 1) / 2;
    Link(paru, parv);
  }
  while (dem <= m) {
    kq[a[dem][1]] = ans;
    dem++;
  }
  for (int i = 1; i <= m; i++) {
    cout << kq[i] << " ";
  }
}