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 chmax(int& a, int b) { a = max(a, b); }
int N, K;
int A[200];
vector<int> edges[200];
int dp[200][202];
void dfs(int i, int p) {
  for (int j : edges[i])
    if (j != p) {
      dfs(j, i);
      int sub[202] = {0};
      for (int a = 0; a <= K + 1; a++)
        for (int b = 0; b <= K + 1; b++)
          if (a + b + 1 > K) {
            chmax(sub[min(a, b + 1)], dp[i][a] + dp[j][b]);
          }
      for (int a = 0; a <= K + 1; a++) chmax(dp[i][a], sub[a]);
    }
  chmax(dp[i][0], dp[i][K + 1] + A[i]);
}
int main() {
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> A[i];
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    edges[a - 1].push_back(b - 1);
    edges[b - 1].push_back(a - 1);
  }
  dfs(0, -1);
  int ans = *max_element(dp[0], dp[0] + K + 2);
  cout << ans << endl;
  return 0;
}