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 c = 202;
int n, m, t[c], dp[c][c];
vector<int> sz[c];
bool v[c];
void dfs(int a) {
  v[a] = true;
  int f = -1;
  for (int i = 0; i < sz[a].size(); i++) {
    int x = sz[a][i];
    if (!v[x]) {
      dfs(x);
      dp[a][0] += dp[x][m];
    } else
      f = i;
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < sz[a].size(); j++) {
      if (f != j) {
        int sum = dp[sz[a][j]][i - 1];
        for (int k = 0; k < sz[a].size(); k++) {
          if (k != f && k != j) sum += dp[sz[a][k]][max(i - 1, m - i)];
        }
        dp[a][i] = max(dp[a][i], sum);
      }
    }
  }
  for (int i = m; i >= 0; i--) {
    dp[a][i] = max(dp[a][i], dp[a][i + 1]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> dp[i][0];
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    sz[a].push_back(b), sz[b].push_back(a);
  }
  dfs(1);
  cout << dp[1][0];
  return 0;
}