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 = 3e5 + 500;
const long long mod = 1e9 + 7;
const long long INF = 1LL << 57;
const int M = 1e6 + 500;
long long int res = 0;
long long int w[N];
long long int ans[N];
long long int get(long long int x) {
  if (x == w[x]) return x;
  return w[x] = get(w[x]);
}
void merge(long long int a, long long int b) {
  a = get(a);
  b = get(b);
  if (a == b) return;
  long long int t = ans[a];
  res -= (t * (t - 1)) / 2;
  t = ans[b];
  res -= (t * (t - 1)) / 2;
  ans[a] += ans[b];
  t = ans[a];
  res += (t * (t - 1)) / 2;
  w[b] = a;
}
vector<pair<long long int, pair<long long int, long long int>>> graph;
long long int p[N];
void solve() {
  for (int i = 0; i < N; i++) {
    ans[i] = 1;
    w[i] = i;
  }
  long long int n, m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    long long int u, v, w;
    cin >> u >> v >> w;
    graph.push_back({w, {u, v}});
  }
  long long int q[m], x[m];
  for (int i = 0; i < m; i++) {
    cin >> q[i];
  }
  sort(graph.begin(), graph.end());
  long long int prev = 0;
  for (auto it : graph) {
    long long int u = it.second.first;
    long long int v = it.second.second;
    merge(u, v);
    p[it.first] = res;
  }
  for (int i = 1; i < N; i++) {
    if (p[i] == 0) p[i] = p[i - 1];
  }
  for (int i = 0; i < m; i++) {
    cout << p[q[i]] << ' ';
  }
  cout << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}