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 = 2e2 + 9;
const int mod = 1e9 + 7;
int n, k;
int a[N], u, v, memo[N][N];
vector<int> adj[N];
int dp(int s, int ht, int p = -1) {
  ht = max(0, ht);
  if (ht > k) return 0;
  if (memo[s][ht] != -1) return memo[s][ht];
  int ret = 0;
  if (ht == 0) {
    ret = a[s];
    for (auto u : adj[s])
      if (u != p) ret += dp(u, k - 1, s);
  } else {
    int sum = 0;
    for (auto u : adj[s])
      if (u != p) sum += dp(u, max(ht - 1, k - ht - 1), s);
    for (auto u : adj[s])
      if (u != p) {
        int tmp = sum + dp(u, ht - 1, s) - dp(u, max(ht - 1, k - ht - 1), s);
        ret = max(ret, tmp);
      }
  }
  return memo[s][ht] = max(ret, dp(s, ht + 1, p));
}
void solve() {
  memset(memo, -1, sizeof(memo));
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  k++;
  cout << dp(0, 0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}