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;
int64_t n, k;
vector<vector<int64_t>> dp;
vector<int64_t> w;
vector<vector<int64_t>> tree;
void calc(int64_t t, int64_t p = -1) {
  if (p != -1 && tree[t].size() == 1) {
    dp[t][0] = w[t];
    return;
  }
  for (int64_t i = 0; i < tree[t].size(); ++i) {
    int64_t e = tree[t][i];
    if (e == p) {
      continue;
    }
    calc(e, t);
  }
  dp[t][0] = w[t];
  for (int64_t i = 0; i < tree[t].size(); ++i) {
    int64_t e = tree[t][i];
    if (e == p) {
      continue;
    }
    dp[t][0] += dp[e][k - 1];
  }
  for (int64_t dep = 1; dep < k; ++dep) {
    for (int64_t i = 0; i < tree[t].size(); ++i) {
      int64_t e = tree[t][i];
      if (e == p) {
        continue;
      }
      int64_t sum = dp[e][dep - 1];
      for (int64_t j = 0; j < tree[t].size(); ++j) {
        int64_t f = tree[t][j];
        if (f == e || f == p) {
          continue;
        }
        sum += dp[f][max(dep - 1, k - dep - 1)];
      }
      dp[t][dep] = max(dp[t][dep], sum);
    }
  }
  for (int64_t dep = k - 1; dep > 0; --dep) {
    dp[t][dep - 1] = max(dp[t][dep - 1], dp[t][dep]);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  ++k;
  dp.assign(n, vector<int64_t>(k));
  w.resize(n);
  tree.resize(n);
  for (int64_t i = 0; i < n; ++i) {
    cin >> w[i];
  }
  for (int64_t i = 0; i < n - 1; ++i) {
    int64_t a, b;
    cin >> a >> b;
    --a;
    --b;
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  calc(0);
  cout << dp[0][0];
}