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 long long mod = 1e9 + 7;
const int maxn = 205;
int n, k, a[maxn];
int dp[maxn][maxn];
vector<int> load[maxn];
void dfs(int s, int pre) {
  dp[s][0] = a[s];
  for (int &e : load[s]) {
    if (e == pre) continue;
    dfs(e, s);
    dp[s][0] += dp[e][k];
  }
  for (int i = 1; i <= k; i++) {
    for (int &e : load[s]) {
      if (e == pre) continue;
      int sum = dp[e][i - 1];
      for (int &u : load[s]) {
        if (u == pre || u == e) continue;
        sum = sum + dp[u][max(i - 1, k - i)];
      }
      dp[s][i] = max(dp[s][i], sum);
    }
  }
  for (int i = k; i >= 1; i--) dp[s][i - 1] = max(dp[s][i - 1], dp[s][i]);
}
int main() {
  int s, e;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    cin >> s >> e;
    load[s].push_back(e);
    load[e].push_back(s);
  }
  dfs(1, -1);
  cout << dp[1][0] << endl;
  return 0;
}