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 main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.setf(ios::fixed);
  cout.precision(15);
  int n, k;
  while (cin >> n >> k) {
    vector<int> w(n);
    for (int i = 0; i < n; i++) {
      cin >> w[i];
    }
    vector<vector<int>> g(n);
    for (int i = 0; i < n - 1; i++) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    if (n == 1) {
      cout << w[0] << endl;
      continue;
    }
    vector<vector<int>> dp(n, vector<int>(n));
    function<void(int, int)> dfs = [&](int x, int p) {
      for (int y : g[x]) {
        if (y == p) continue;
        dfs(y, x);
      }
      for (int y : g[x]) {
        if (y == p) continue;
        for (int hy = 0; hy < n; hy++) {
          int test = dp[y][hy];
          for (int z : g[x]) {
            if (z == p) continue;
            if (z == y) continue;
            int hz = max(hy, k - hy - 1);
            if (hz >= n) continue;
            assert(hy + hz + 2 > k);
            test += dp[z][hz];
          }
          if (hy + 1 < n) {
            dp[x][hy + 1] = max(dp[x][hy + 1], test);
          }
        }
      }
      dp[x][0] = w[x];
      for (int y : g[x]) {
        if (y == p) continue;
        if (k < n) dp[x][0] += dp[y][k];
      }
      for (int h = n - 2; h >= 0; h--) {
        dp[x][h] = max(dp[x][h], dp[x][h + 1]);
      }
    };
    dfs(0, -1);
    int ans = 0;
    for (int h = 0; h < n; h++) {
      ans = max(ans, dp[0][h]);
    }
    cout << ans << endl;
  }
}