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;
vector<long long> r1, sz, p;
void mk(long long v) {
  sz[v] = 1;
  r1[v] = 1;
  p[v] = v;
}
long long get(long long v) {
  if (v == p[v]) return v;
  return p[v] = get(p[v]);
}
long long ans = 0;
void unite(long long a, long long b) {
  a = get(a), b = get(b);
  if (a == b) return;
  if (r1[b] > r1[a]) swap(a, b);
  if (r1[a] == r1[b]) r1[a]++;
  ans -= sz[a] * (sz[a] - 1) / 2;
  ans -= sz[b] * (sz[b] - 1) / 2;
  sz[a] += sz[b];
  p[b] = a;
  ans += sz[a] * (sz[a] - 1) / 2;
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second < b.second;
}
signed main() {
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, pair<long long, long long>>> ed;
  r1.resize(n), sz.resize(n), p.resize(n);
  for (long long i = 0; i < n - 1; ++i) {
    long long u, v, c;
    cin >> u >> v >> c;
    u--, v--;
    ed.push_back({c, {u, v}});
  }
  sort(ed.begin(), ed.end());
  vector<long long> res(m);
  vector<pair<long long, long long>> q(m);
  for (long long i = 0; i < m; ++i) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  for (long long i = 0; i < n; ++i) mk(i);
  long long j = -1;
  for (long long i = 0; i < m; ++i) {
    while (j + 1 < n - 1 && ed[j + 1].first <= q[i].first) {
      unite(ed[j + 1].second.first, ed[j + 1].second.second), j++;
    }
    res[q[i].second] = ans;
  }
  for (long long i = 0; i < m; ++i) cout << res[i] << ' ';
  return 0;
}