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;
int n, k;
int v[300], maxDepth[300];
int s[300][300];
int depth[300];
vector<int> g[300];
int r;
void f(int n) {
  maxDepth[n] = 0;
  for (int c : g[n])
    if (depth[c] == 0) {
      depth[c] = depth[n] + 1;
      f(c);
      maxDepth[n] = max(maxDepth[c] + 1, maxDepth[n]);
    }
  int prev = 0;
  for (int d = min(k, maxDepth[n]); d > 0; --d) {
    if (d * 2 > k) {
      for (int c : g[n])
        if (depth[c] > depth[n]) s[n][d] += s[c][d - 1];
    } else
      for (int c1 : g[n])
        if (depth[c1] > depth[n]) {
          int w = s[c1][d - 1];
          for (int c2 : g[n]) {
            if (c1 != c2 && depth[c2] > depth[n]) w += s[c2][k - d];
          }
          s[n][d] = max(s[n][d], w);
        }
    s[n][d] = max(s[n][d], prev);
    prev = s[n][d];
  }
  s[n][0] = v[n];
  for (int c : g[n]) {
    if (depth[c] > depth[n]) s[n][0] += s[c][k];
  }
  s[n][0] = max(s[n][0], prev);
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> v[i + 1];
  int start = 1;
  int end = 0;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
    start = min(a, start);
    start = min(b, start);
    end = min(a, end);
    end = max(a, end);
  }
  depth[1] = 1;
  f(1);
  int ans = 0;
  for (int i = 1; i <= n; ++i) ans = max(s[i][0], ans);
  cout << ans << endl;
}