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 maxn = 200 + 10;
vector<int> G[maxn];
int dp[maxn][maxn];
int weight[maxn];
int n, k;
void dfs(int now, int fa) {
  dp[now][0] = weight[now];
  for (auto it : G[now]) {
    if (it == fa) continue;
    dfs(it, now);
  }
  for (int dep = 0; dep <= n; ++dep) {
    if (dep == 0) {
      for (auto it : G[now]) {
        if (it == fa) continue;
        dp[now][0] += dp[it][max(0, k - 1)];
      }
    } else {
      for (auto son : G[now]) {
        if (son == fa) continue;
        int cnt = dp[son][dep - 1];
        for (auto other : G[now]) {
          if (other == fa || other == son) continue;
          cnt += dp[other][max(dep - 1, k - dep - 1)];
        }
        dp[now][dep] = max(dp[now][dep], cnt);
      }
    }
  }
  for (int i = n - 1; i >= 0; --i) dp[now][i] = max(dp[now][i], dp[now][i + 1]);
}
int main() {
  cin >> n >> k;
  k++;
  for (int i = 1; i <= n; ++i) cin >> weight[i];
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, 0);
  printf("%d\n", dp[1][0]);
  return 0;
}