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;
long long n, k, dp[209][209], c[500009], ans, cur;
vector<long long> v[500009];
void go(long long x, long long parent) {
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] == parent) continue;
    go(v[x][i], x);
  }
  for (long long first = 1; first <= n; first++) {
    for (int i = 0; i < v[x].size(); i++) {
      if (v[x][i] == parent) continue;
      cur = 0;
      cur += dp[v[x][i]][first - 1];
      for (int j = 0; j < v[x].size(); j++) {
        if (v[x][j] != parent && i != j)
          cur += dp[v[x][j]][max(first - 1, k - first - 1LL)];
      }
      if (dp[x][first] < cur) dp[x][first] = cur;
    }
  }
  dp[x][0] += c[x];
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != parent) {
      dp[x][0] += dp[v[x][i]][k - 1] + cur;
      dp[x][0] -= cur;
    }
  for (int i = n; i >= 0; i--) {
    if (dp[x][i] < dp[x][i + 1]) dp[x][i] = dp[x][i + 1];
  }
  ans = max(ans, dp[x][0]);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  k++;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    v[y].push_back(x);
    v[x].push_back(y);
  }
  go(1, 0);
  cout << ans << endl;
  return 0;
}