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;
using ll = long long;
using ldb = long double;
using pa = pair<ll, ll>;
using vec = vector<ll>;
const int MOD = 1e9 + 7, INF = INT_MAX, N = 1e5 + 10;
const double PI = acos(-1);
const ll LINF = LLONG_MAX;
ll parent[200009], sizes[200009];
void make_set(int v) {
  parent[v] = v;
  sizes[v] = 1;
}
int find_set(int v) {
  if (v == parent[v]) return v;
  return parent[v] = find_set(parent[v]);
}
ll re = 0;
ll cnt(ll a) { return 1ll * a * (a - 1) / 2; }
void union_sets(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (sizes[a] < sizes[b]) swap(a, b);
    re -= cnt(sizes[a]);
    re -= cnt(sizes[b]);
    parent[b] = a;
    sizes[a] += sizes[b];
    re += cnt(sizes[a]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll n, m;
  cin >> n >> m;
  vector<array<int, 3>> inputs(n - 1);
  for (ll i = 0; i < n - 1; i++) {
    cin >> inputs[i][1] >> inputs[i][2] >> inputs[i][0];
    inputs[i][1]--;
    inputs[i][2]--;
  }
  vector<pa> weights(m);
  vec answer(m);
  for (ll i = 0; i < m; i++) {
    cin >> weights[i].first;
    weights[i].second = i;
  }
  for (ll i = 0; i < n + 1; i++) {
    make_set(i);
  }
  sort(inputs.begin(), inputs.end());
  sort(weights.begin(), weights.end());
  ll pos = 0;
  for (ll i = 0; i < m; i++) {
    while (pos < n - 1 && inputs[pos][0] <= weights[i].first) {
      union_sets(inputs[pos][1], inputs[pos][2]);
      ++pos;
    }
    answer[weights[i].second] = re;
  }
  for (int i = 0; i < m; ++i) {
    cout << answer[i] << " ";
  }
  cerr << "Time taken: " << int((clock() * 1000.) / CLOCKS_PER_SEC) << "ms\n";
  return 0;
}