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 = 206;
vector<int> E[N];
int dp[N][N], c[N];
int n, k;
void dfs(int node, int p) {
  for (auto it : E[node]) {
    if (it != p) {
      dfs(it, node);
    }
  }
  for (int d = 0; d <= N - 1; d += 1) {
    if (d == 0) {
      dp[node][d] = c[node];
      for (auto it : E[node]) {
        if (it != p) {
          dp[node][d] += dp[it][max(0, k - d - 1)];
        }
      }
    }
    if (d > 0) {
      for (auto it : E[node]) {
        if (it != p) {
          int cur = dp[it][d - 1];
          for (auto itt : E[node]) {
            if (itt != it && itt != p) {
              cur += dp[itt][max(d - 1, k - d - 1)];
            }
          }
          dp[node][d] = max(dp[node][d], cur);
        }
      }
    }
  }
  for (int i = N - 2; i >= 0; i--) {
    dp[node][i] = max(dp[node][i], dp[node][i + 1]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  k += 1;
  for (int i = 1; i <= n; i++) cin >> c[i];
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    cin >> x >> y;
    E[x].push_back(y);
    E[y].push_back(x);
  }
  dfs(1, 0);
  cout << dp[1][0] << '\n';
  return 0;
}