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 = 201;
int n, k;
int w[N];
vector<int> g[N];
int dp[N][N];
int cur[N];
void dfs(int v, int last = -1) {
  for (int u : g[v]) {
    if (u != last) {
      dfs(u, v);
    }
  }
  fill(cur, cur + N, 0);
  for (int u : g[v]) {
    if (u != last) {
      for (int h = 0; h <= k; ++h) {
        cur[h] += dp[u][h];
      }
    }
  }
  for (int d = 0; d <= k; ++d) {
    for (int m = max(0, d - 1); m < k && max(k - 1 - m, d - 1) >= m; ++m) {
      for (int u : g[v]) {
        if (u != last) {
          dp[v][d] = max(dp[v][d], cur[max(k - 1 - m, d - 1)] -
                                       dp[u][max(k - 1 - m, d - 1)] + dp[u][m]);
        }
      }
    }
  }
  int a = w[v];
  for (int u : g[v]) {
    if (u != last) {
      a += dp[u][k];
    }
  }
  dp[v][0] = max(dp[v][0], a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  for (int v = 0; v < n; ++v) {
    cin >> w[v];
  }
  for (int i = 0; i < n - 1; ++i) {
    int v, u;
    cin >> v >> u;
    --v;
    --u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0);
  cout << dp[0][0] << '\n';
  return 0;
}