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;
long long int _gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return _gcd(b, a % b);
}
long long int par[200005], sz[200005];
long long int fin(long long int x) {
  if (par[x] == x) return x;
  return fin(par[x]);
}
void unio(long long int x, long long int y) {
  long long int p1 = fin(x);
  long long int p2 = fin(y);
  if (sz[p1] >= sz[p2]) {
    sz[p1] += sz[p2];
    par[p2] = p1;
  } else {
    sz[p2] += sz[p1];
    par[p1] = p2;
  }
}
void dsu(long long int n) {
  for (long long int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j = 0, res = 0;
  cin >> n >> m;
  vector<pair<long long int, pair<long long int, long long int>>> edge;
  vector<pair<long long int, long long int>> query;
  long long int ans[m];
  dsu(n);
  for (i = 0; i < n - 1; i++) {
    long long int x, y, w;
    cin >> x >> y >> w;
    edge.push_back(make_pair(w, make_pair(x, y)));
  }
  for (i = 0; i < m; i++) {
    long long int x;
    cin >> x;
    query.push_back(make_pair(x, i));
  }
  sort(edge.begin(), edge.end());
  sort(query.begin(), query.end());
  for (i = 0; i < m; i++) {
    while (j < n - 1 && edge[j].first <= query[i].first) {
      long long int p1 = fin(edge[j].second.first);
      long long int p2 = fin(edge[j].second.second);
      res -= ((sz[p1] * (sz[p1] - 1)) / 2) + ((sz[p2] * (sz[p2] - 1)) / 2) -
             (((sz[p1] + sz[p2]) * (sz[p1] + sz[p2] - 1)) / 2);
      j++;
      unio(p1, p2);
    }
    ans[query[i].second] = res;
  }
  for (auto x : ans) cout << x << ' ';
  return 0;
}