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 N = 200 + 5;
int n, k, a[N], dp[N][N];
vector<int> g[N];
void dfs(int u, int pre) {
  dp[u][0] = a[u];
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == pre) continue;
    dfs(v, u);
  }
  for (int d = 0; d < n; d++) {
    if (!d) {
      for (int i = 0; i < g[u].size(); i++) {
        if (g[u][i] == pre) continue;
        dp[u][d] += dp[g[u][i]][max(0, k - d - 1)];
      }
    } else {
      for (int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if (v == pre) continue;
        int ans = dp[v][d - 1];
        for (int j = 0; j < g[u].size(); j++) {
          int v1 = g[u][j];
          if (v1 == pre || v1 == v) continue;
          ans += dp[v1][max(d - 1, k - d - 1)];
        }
        dp[u][d] = max(dp[u][d], ans);
      }
    }
  }
  for (int i = n - 1; i >= 0; i--) dp[u][i] = max(dp[u][i], dp[u][i + 1]);
}
int main() {
  cin >> n >> k;
  ++k;
  for (int i = 1; i <= n; i++) cin >> a[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 << dp[1][0] << endl;
  return 0;
}