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;
auto IOS = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
const int MAXN = 222;
int n, k, w[MAXN], f[MAXN][MAXN];
vector<int> G[MAXN];
void dfs(int u, int fa) {
  f[u][0] = w[u];
  if (G[u].size() == 1 && fa != 0) return;
  for (int v : G[u]) {
    if (v == fa) continue;
    dfs(v, u);
    f[u][0] += f[v][k];
  }
  for (int to : G[u]) {
    if (to == fa) continue;
    for (int dep = 0; dep <= k; dep++) {
      int tp = f[to][dep];
      for (int v : G[u]) {
        if (v == fa || v == to) continue;
        tp += f[v][max(dep, k - dep - 1)];
      }
      f[u][dep + 1] = max(f[u][dep + 1], tp);
    }
  }
  for (int dep = k - 1; dep >= 0; dep--)
    f[u][dep] = max(f[u][dep], f[u][dep + 1]);
}
int main() {
  IOS();
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> w[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  cout << f[1][0] << endl;
  return 0;
}