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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using db = long double;
using ii = pair<int, int>;
const int N = 2e5 + 5, LG = 17, MOD = 1e9 + 7;
const int SQ = 450;
const long double EPS = 1e-7;
int dp[205][205][205];
int a[205];
int n, k;
vector<int> adj[205];
int solve(int node, int par, int d) {
  d = max(d, 0);
  int &ret = dp[node][par][d];
  if (~ret) return ret;
  ret = 0;
  if (d == 0) {
    int choice = 0;
    for (auto v : adj[node])
      if (v != par) {
        choice += solve(v, node, k);
      }
    ret = max(ret, choice + a[node]);
  }
  for (int mn = max(0, d - 1); mn <= k; mn++) {
    int mx = max(d - 1, k - mn - 1);
    if (mx < mn) break;
    int sum1 = 0;
    for (auto v : adj[node])
      if (v != par) {
        sum1 += solve(v, node, mx);
      }
    for (auto v : adj[node])
      if (v != par) {
        ret = max(ret, sum1 - solve(v, node, mx) + solve(v, node, mn));
      }
  }
  return ret;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i < n + 1; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  memset(dp, -1, sizeof dp);
  int ans = 0;
  for (int i = 1; i < n + 1; i++) {
    ans = max(ans, solve(i, i, k + 1) + a[i]);
  }
  cout << ans << '\n';
  return 0;
}