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 MAXN = 200 + 17, inf = 2e9 + 17;
int n, k, a[MAXN], dp[MAXN][MAXN], par[MAXN];
vector<int> adj[MAXN];
void dfs(int v) {
  if (adj[v].size() == 1 && v != 1) {
    dp[v][0] = a[v];
    return;
  }
  for (auto u : adj[v])
    if (u != par[v]) {
      par[u] = v;
      dfs(u);
    }
  int sum[MAXN] = {};
  for (int i = 0; i <= k; i++)
    for (auto u : adj[v])
      if (u != par[v]) sum[i] += dp[u][i];
  dp[v][0] = sum[k] + a[v];
  for (int i = k; i > 0; i--) {
    int res = dp[v][i + 1];
    for (auto u : adj[v])
      if (u != par[v]) {
        int tmp = max(i, k - i + 1);
        res = max(res, sum[tmp - 1] - dp[u][tmp - 1] + dp[u][i - 1]);
      }
    dp[v][i] = res;
  }
  dp[v][0] = max(dp[v][0], dp[v][1]);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  dfs(1);
  cout << dp[1][0] << endl;
  return 0;
}