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;
using ll = long long;
int const nmax = 200;
vector<int> g[5 + nmax];
int val[5 + nmax];
int dp[5 + nmax][5 + nmax];
int dp2[5 + nmax][5 + nmax];
int n, k;
void dfs(int node, int parent) {
  dp[node][0] = val[node];
  for (int h = 0; h < g[node].size(); h++) {
    int to = g[node][h];
    if (to != parent) {
      dfs(to, node);
      for (int i = 0; i <= nmax; i++) dp2[node][i] = 0;
      for (int i = 0; i <= nmax; i++)
        for (int j = 0; j < nmax; j++)
          if (k < i + j + 1)
            dp2[node][min(i, j + 1)] =
                max(dp2[node][min(i, j + 1)], dp[node][i] + dp[to][j]);
      for (int i = 0; i < nmax; i++)
        dp[node][i] = max(dp[node][i], dp2[node][i]);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> val[i];
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  int result = 0;
  for (int i = 0; i <= nmax; i++) result = max(result, dp[1][i]);
  cout << result;
  return 0;
}