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 long long M = 1e9 + 7;
const long long N = 2e5;
long long po(long long, long long);
long long n, m;
long long par[N + 5];
long long sz[N + 5];
long long ans;
long long get(long long x) {
  if (par[x] == x)
    return x;
  else
    return par[x] = get(par[x]);
}
void merge(long long a, long long b) {
  a = get(a);
  b = get(b);
  if (sz[a] > sz[b]) {
    swap(a, b);
  }
  ans -= (sz[a] * (sz[a] - 1) / 2);
  ans -= (sz[b] * (sz[b] - 1) / 2);
  sz[b] += sz[a];
  ans += (sz[b] * (sz[b] - 1) / 2);
  par[a] = b;
}
void solve() {
  cin >> n >> m;
  for (long long i = 1; i < n + 1; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  vector<pair<long long, pair<long long, long long>>> edge;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y, w;
    cin >> x >> y >> w;
    edge.push_back({w, {x, y}});
  }
  vector<pair<long long, long long>> Q;
  for (long long i = 0; i < m; i++) {
    long long x;
    cin >> x;
    Q.push_back({x, i});
  }
  sort((Q).begin(), (Q).end());
  sort((edge).begin(), (edge).end());
  ans = 0;
  long long res[m + 1];
  long long j = 0;
  for (long long i = 0; i < m; i++) {
    long long q = Q[i].first;
    long long id = Q[i].second;
    while (j < n - 1 && edge[j].first <= q) {
      merge(edge[j].second.first, edge[j].second.second);
      j++;
    }
    res[id] = ans;
  }
  for (long long i = 0; i < m; i++) {
    cout << res[i] << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 1; i < t + 1; i++) {
    solve();
  }
  return 0;
}
long long po(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = 1;
  if (b % 2 == 0) {
    ans = po(a, b / 2) % M;
    ans = (ans % M * ans % M) % M;
  } else {
    ans = po(a, (b - 1) / 2) % M;
    ans = (ans % M * ans % M * a) % M;
  }
  return ans % M;
}