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;
typedef struct node {
  long long int val, more = 0, less = 0;
  bool operator<(const node& a) { return val < a.val; }
} node;
const int M = 1e9 + 7;
int par[200005];
int find_par(int v) {
  if (par[v] < 0) return v;
  return par[v] = find_par(par[v]);
}
long long int dsu(int x, int y) {
  x = find_par(x);
  y = find_par(y);
  long long int ans = (long long int)par[x] * par[y];
  if (par[y] < par[x]) swap(x, y);
  par[x] += par[y];
  par[y] = x;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  memset(par, -1, sizeof(par));
  vector<pair<int, pair<int, int>>> edge;
  for (int i = 1, x, y, w; i < n;
       ++i, cin >> x >> y >> w, edge.push_back({w, {x, y}})) {
  }
  sort(edge.begin(), edge.end());
  vector<pair<int, long long int>> max_pair;
  max_pair.push_back({0, 0});
  long long int cnt = 0;
  for (int i = 0; i < n - 1; ++i) {
    cnt += dsu(edge[i].second.first, edge[i].second.second);
    if (i == n - 2 || edge[i].first != edge[i + 1].first)
      max_pair.push_back({edge[i].first, cnt});
  }
  for (int i = 0, q; i < m; ++i) {
    cin >> q;
    std::vector<pair<int, long long int>>::iterator low =
        lower_bound(max_pair.begin(), max_pair.end(), make_pair(q, LLONG_MAX));
    if (low == max_pair.end())
      cout << (long long int)n * (n - 1) / 2 << " ";
    else {
      cout << (--low)->second << " ";
    }
  }
  return 0;
}