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 max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int min(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
long long int powmd(long long int a, long long int b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long int half = powmd(a, b / 2);
  long long int ans = half * half;
  ans %= 1000000007;
  if (b & 1) {
    ans *= a;
  }
  ans %= 1000000007;
  return ans;
}
void fastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
struct edge {
  long long int a, b, wt;
  bool operator<(const edge& e) const { return wt < e.wt; }
};
edge edges[200005];
long long int parent[200005];
long long int sizes[200005];
struct dsu {
  void init(long long int n) {
    for (long long int i = 0; i < n; i++) {
      parent[i] = i;
      sizes[i] = 1;
    }
  }
  long long int get_parent(long long int x) {
    if (parent[x] == x) return x;
    return parent[x] = get_parent(parent[x]);
  }
  long long int _union(long long int a, long long int b) {
    long long int a_parent = get_parent(a);
    long long int b_parent = get_parent(b);
    long long int ans = sizes[b_parent] * sizes[a_parent];
    sizes[b_parent] += sizes[a_parent];
    parent[a_parent] = b_parent;
    return ans;
  }
};
int32_t main() {
  fastIO();
  long long int t = 1;
  while (t--) {
    long long int n, q;
    cin >> n >> q;
    dsu graph;
    graph.init(n);
    long long int a, b, wt;
    for (long long int i = 0; i < n - 1; i++) {
      cin >> a >> b >> wt;
      a--, b--;
      edges[i].a = a;
      edges[i].b = b;
      edges[i].wt = wt;
    }
    sort(edges, edges + n - 1);
    map<long long int, long long int> m;
    for (long long int i = 0; i < n - 1; i++) {
      m[edges[i].wt] += graph._union(edges[i].a, edges[i].b);
    }
    vector<pair<long long int, long long int> > prefix(m.size());
    long long int sum = 0;
    for (auto it : m) {
      sum += it.second;
      prefix.push_back(make_pair(it.first, sum));
    }
    long long int query;
    pair<long long int, long long int> curr_pair;
    for (long long int i = 0; i < q; i++) {
      cin >> query;
      curr_pair.first = query;
      curr_pair.second = 1e18;
      auto it = upper_bound(prefix.begin(), prefix.end(), curr_pair);
      if (it == prefix.begin()) {
        cout << 0 << " ";
      } else {
        it--;
        cout << it->second << " ";
      }
    }
    cout << endl;
    ;
  }
  return 0;
}