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;
long long int un = 0;
long long int valr(long long int x) { return (x * (x - 1) * 1LL) / 2; }
int find(int x, int par[]) {
  if (par[x] == x)
    return x;
  else {
    int zx = find(par[x], par);
    par[x] = zx;
    return zx;
  }
}
void uni(int x, int y, int par[], int rank[], long long int weight[]) {
  find(x, par);
  find(y, par);
  if (par[x] == par[y]) return;
  if (rank[par[x]] == rank[par[y]]) {
    un = un - valr(weight[par[x]]) - valr(weight[par[y]]);
    weight[par[x]] = weight[par[x]] + weight[par[y]];
    weight[par[y]] = 0;
    un = un + valr(weight[par[x]]);
    par[par[y]] = par[x];
    rank[par[x]]++;
  } else if (rank[par[x]] > rank[par[y]]) {
    un = un - valr(weight[par[x]]) - valr(weight[par[y]]);
    weight[par[x]] = weight[par[x]] + weight[par[y]];
    weight[par[y]] = 0;
    un = un + valr(weight[par[x]]);
    par[par[y]] = par[x];
  } else {
    un = un - valr(weight[par[x]]) - valr(weight[par[y]]);
    weight[par[y]] = weight[par[y]] + weight[par[x]];
    weight[par[x]] = 0;
    un = un + valr(weight[par[y]]);
    par[par[x]] = par[y];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  if (n == 1) {
    int xxc;
    for (int i = 0; i < m; i++) cin >> xxc;
    for (int i = 0; i < m; i++) cout << 0 << " ";
    exit(0);
  }
  pair<int, pair<int, int>> p[n - 1];
  int u, v, w;
  int ma = -1;
  for (int i = 0; i < n - 1; i++) {
    cin >> p[i].second.first >> p[i].second.second >> p[i].first;
    if (p[i].first > ma) ma = p[i].first;
  }
  int par[n + 1], rank[n + 1];
  long long int weight[n + 1];
  for (int i = 0; i <= n; i++) {
    par[i] = i;
    rank[i] = 1;
    weight[i] = 1;
  }
  long long int ans[ma + 1];
  sort(p, p + n - 1);
  ans[0] = 0;
  int j = 0;
  for (int i = 1; i <= ma; i++) {
    if (p[j].first > i || j >= n - 1) {
      ans[i] = ans[i - 1];
    } else {
      while (p[j].first <= i && j < n - 1) {
        uni(p[j].second.first, p[j].second.second, par, rank, weight);
        j++;
      }
      ans[i] = un;
    }
  }
  int pp;
  for (int i = 0; i < m; i++) {
    cin >> pp;
    if (pp > ma)
      cout << ans[ma] << " ";
    else
      cout << ans[pp] << " ";
  }
  return 0;
}