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;
void dfs(vector<int64_t>* edges, int64_t* wt, int64_t** dp, int64_t n,
         int64_t k, int64_t v, int64_t p) {
  for (int64_t i = 0; i < edges[v].size(); i++) {
    int64_t c = edges[v][i];
    if (c != p) {
      dfs(edges, wt, dp, n, k, c, v);
    }
  }
  dp[v][0] = wt[v];
  for (int64_t d = 0; d < 202; d++) {
    for (int64_t i = 0; i < edges[v].size(); i++) {
      int64_t c = edges[v][i];
      int64_t cur = 0;
      if (d == 0)
        cur = wt[v];
      else
        cur = dp[c][d - 1];
      if (c != p) {
        if (d == 0) {
          dp[v][0] += dp[c][max((int64_t)0, k - 1)];
        } else {
          for (int64_t i = 0; i < edges[v].size(); i++)
            if (edges[v][i] != p && edges[v][i] != c) {
              int64_t c2 = edges[v][i];
              cur += dp[c2][max(d - 1, k - d - 1)];
            }
        }
      }
      dp[v][d] = max(dp[v][d], cur);
    }
  }
  for (int64_t d = 202 - 2; d >= 0; d--) {
    dp[v][d] = max(dp[v][d], dp[v][d + 1]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int64_t n, k;
  cin >> n >> k;
  ++k;
  int64_t* wt = new int64_t[n];
  for (int64_t i = 0; i < n; i++) {
    cin >> wt[i];
  }
  vector<int64_t>* edges = new vector<int64_t>[n];
  for (int64_t i = 0; i < n - 1; i++) {
    int64_t x, y;
    cin >> x >> y;
    --x;
    --y;
    edges[x].push_back(y);
    edges[y].push_back(x);
  }
  int64_t** dp = new int64_t*[n];
  for (int64_t i = 0; i < n; i++) {
    dp[i] = new int64_t[202]();
  }
  dfs(edges, wt, dp, n, k, 0, -1);
  cout << dp[0][0] << endl;
  delete[] wt;
  delete[] edges;
  for (int64_t i = 0; i < n; i++) delete[] dp[i];
  delete[] dp;
  return 0;
}