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 N = 200005;
const long long INF = 1e9;
const long long MOD = 1000000007;
int n, k;
int arr[203];
vector<vector<int>> ve(203);
vector<int> dfs(int ind, int prev) {
  vector<int> v1(k + 2, 0);
  v1[0] = arr[ind];
  for (auto i : ve[ind]) {
    if (i == prev) continue;
    vector<int> v2;
    v2 = dfs(i, ind);
    vector<int> v3(k + 3, 0);
    for (int i = 1; i <= k + 1; i++) {
      if (v2[i - 1] != 0) {
        for (int j = k + 1 - i; j <= k + 1; j++)
          if (v1[j] != 0) v3[min(i, j)] = max(v3[min(i, j)], v2[i - 1] + v1[j]);
      }
      v3[i] = max(v3[i], v2[i - 1]);
    }
    for (int i = 0; i <= k; i++) v1[i] = max(v1[i], v3[i]);
    for (int i = 0; i <= k; i++)
      for (int j = i + 1; j <= k + 1; j++) v1[i] = max(v1[i], v1[j]);
  }
  return v1;
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    ve[x].push_back(y);
    ve[y].push_back(x);
  }
  vector<int> v1 = dfs(1, -1);
  int mm = 0;
  for (auto i : v1) mm = max(i, mm);
  cout << mm;
}