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 = 2e2;
int k, vl[N + 2], dp[N + 2][N + 2];
vector<int> adj[N + 2];
int dfs(int node, int par, int d) {
  if (dp[node][d] != -1) return dp[node][d];
  int now = 0;
  if (!d) {
    for (auto x : adj[node])
      if (x != par) now += dfs(x, node, k);
    now += vl[node];
  }
  vector<vector<int> > ad, qm;
  ad.resize(k + 1), qm.resize(k + 1);
  for (int i = 0; i <= k; i++) {
    ad[i].resize((int)adj[node].size() + 1);
    qm[i].resize((int)adj[node].size() + 1);
  }
  int id = 0;
  for (auto x : adj[node]) {
    if (x == par) continue;
    id++;
    for (int i = 0; i <= k; i++) {
      ad[i][id] = dfs(x, node, i);
    }
  }
  for (int i = 0; i <= k; i++)
    for (int j = 1; j <= id; j++) qm[i][j] = qm[i][j - 1] + ad[i][j];
  for (int i = 1; i <= id; i++) {
    for (int j = k; j >= max(0, d - 1); j--) {
      int l = max(0, max(d - 1, max(j, k - 1 - j)));
      now = max(now, ad[j][i] + qm[l][id] - ad[l][i]);
    }
  }
  return dp[node][d] = now;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> vl[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  memset(dp, -1, sizeof(dp));
  cout << dfs(1, 0, 0) << endl;
  return 0;
}