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 = 210;
int A[N], n, K, dp[N][N];
vector<int> adj[N];
bool vis[N];
void read() {
  cin >> n >> K;
  for (int i = 1; i <= n; i++) cin >> A[i];
  for (int i = 1, v, u; i < n; i++) {
    cin >> v >> u;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
}
void dfs(int root, int h = -1) {
  vis[root] = true;
  dp[root][0] += A[root];
  vector<int> vec;
  for (auto u : adj[root])
    if (!vis[u]) {
      vec.push_back(u);
      dfs(u, h);
      dp[root][0] += dp[u][K - 1];
    }
  for (int t = 1; t <= n; t++) {
    int s = 0;
    for (auto u : vec) s += dp[u][max(0, max(K - t - 1, t - 1))];
    for (auto u : vec)
      dp[root][t] = max(
          dp[root][t], dp[u][t - 1] + s - dp[u][max(0, max(K - t - 1, t - 1))]);
  }
  for (int t = n; t >= 0; t--) dp[root][t] = max(dp[root][t], dp[root][t + 1]);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  read();
  K++;
  dfs(1);
  cout << dp[1][0] << endl;
  return 0;
}