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;
template <typename T>
T read1() {
  T t = 0;
  char k;
  bool vis = 0;
  do (k = getchar()) == '-' && (vis = 1);
  while ('0' > k || k > '9');
  while ('0' <= k && k <= '9')
    t = (t << 3) + (t << 1) + (k ^ '0'), k = getchar();
  return vis ? -t : t;
}
int s, k, dp[205][205], a[205];
vector<int> G[205];
void dfs(int n, int fa) {
  dp[n][0] = a[n];
  for (int i = 0; i < G[n].size(); ++i)
    if (G[n][i] ^ fa) {
      dfs(G[n][i], n);
      dp[n][0] += dp[G[n][i]][k];
      for (int j = 1; j <= s; ++j)
        for (int l = j; l <= s; ++l)
          dp[n][j] = max(dp[n][j], dp[n][l] + dp[G[n][i]][max(k - l, j - 1)]);
    }
  for (int i = s; i--;) dp[n][i] = max(dp[n][i], dp[n][i + 1]);
}
int main() {
  s = read1<int>(), k = read1<int>();
  for (int i = 1; i <= s; ++i) a[i] = read1<int>();
  for (int i = 1; i < s; ++i) {
    int u = read1<int>(), v = read1<int>();
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  printf("%d", dp[1][0]);
  return 0;
}