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>
int n, k;
int k2;
int dp[210][210];
int val[210];
std::vector<std::vector<int>> paths(210);
int solve(int i, int left, int p) {
  if (dp[i][left] != -1) return dp[i][left];
  int &ret = dp[i][left];
  ret = 0;
  std::vector<std::vector<int>> dist;
  std::vector<int> sums(k + 1, 0);
  for (int j = 0; j < paths[i].size(); j++) {
    if (paths[i][j] == p) continue;
    dist.push_back(std::vector<int>(k + 1));
    for (int kk = std::max(0, left - 1); kk <= k; kk++) {
      dist.back()[kk] = solve(paths[i][j], kk, i);
      sums[kk] += dist.back()[kk];
    }
  }
  int tmpk = k - 1;
  int tmpk2 = std::max((tmpk) / 2, 0);
  for (int j = 0; j < dist.size(); j++) {
    for (int kk = std::max(left - 1, 0); kk <= tmpk2; kk++) {
      int tmp = std::max(left - 1, tmpk - kk);
      ret = std::max(ret, sums[tmp] + dist[j][kk] - dist[j][tmp]);
    }
  }
  if (left == 0) {
  }
  ret = std::max(ret, sums[std::max(tmpk - tmpk2, left - 1)]);
  if (left == 0) {
    ret = std::max(ret, val[i] + sums[k]);
  }
  return ret;
}
int main() {
  for (int i = 0; i < 210; i++) {
    for (int j = 0; j < 210; j++) {
      dp[i][j] = -1;
    }
  }
  std::cin >> n >> k;
  k2 = k / 2;
  for (int i = 0; i < n; i++) {
    std::cin >> val[i];
  }
  int a, b;
  for (int i = 1; i < n; i++) {
    std::cin >> a >> b;
    paths[--a].push_back(--b);
    paths[b].push_back(a);
  }
  std::cout << solve(0, 0, -1) << "\n";
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
    }
  }
  return 0;
}