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 mod = 1e9 + 7;
long long n, m;
vector<long long> query(200005);
vector<long long> p(200005);
vector<long long> r(200005, 0);
vector<long long> val(200005, 1);
long long cnt;
long long fin(long long i) {
  if (p[i] != i)
    return p[i] = fin(p[i]);
  else
    return p[i];
}
void uni(long long x, long long y) {
  long long xroot = fin(x);
  long long yroot = fin(y);
  if (xroot == yroot) return;
  long long a = val[xroot];
  long long b = val[yroot];
  if (r[xroot] < r[yroot]) {
    p[xroot] = yroot;
    val[yroot] = a + b;
  } else if (r[yroot] < r[xroot]) {
    p[yroot] = xroot;
    val[xroot] = a + b;
  } else {
    p[yroot] = xroot;
    r[xroot]++;
    val[xroot] = a + b;
  }
  cnt -= (a * (a - 1)) / 2 + (b * (b - 1)) / 2;
  long long t = a + b;
  cnt += (t * (t - 1)) / 2;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long>>> v;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y, w;
    cin >> x >> y >> w;
    v.push_back({w, {x, y}});
  }
  for (long long i = 0; i < m; i++) {
    cin >> query[i];
  }
  sort(v.begin(), v.end());
  for (long long i = 0; i <= n; i++) p[i] = i;
  map<long long, long long> ans;
  for (long long i = 0; i < n - 1; i++) {
    uni(v[i].second.first, v[i].second.second);
    ans[v[i].first] = cnt;
  }
  for (long long i = 0; i < m; i++) {
    auto it = ans.upper_bound(query[i]);
    if (it == ans.begin()) {
      cout << 0 << ' ';
      continue;
    }
    it--;
    cout << it->second << ' ';
  }
  return 0;
}