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;
struct query {
  int Q, id;
  long long res;
};
const int Nmax = 200001;
int n, m, parent[Nmax], Rank[Nmax];
vector<pair<int, pair<int, int> > > edge;
bool cmpQ(const query &q1, const query &q2) { return q1.Q < q2.Q; }
bool ret(const query &q1, const query &q2) { return q1.id < q2.id; }
int Find(const int &i) {
  if (i == parent[i]) return i;
  return parent[i] = Find(parent[i]);
}
void Union(const int &i, const int &j) {
  int x = Find(i), y = Find(j);
  if (x != y) {
    if (Rank[x] > Rank[y]) {
      parent[y] = x;
      Rank[x] += Rank[y];
    } else {
      parent[x] = y;
      Rank[y] += Rank[x];
    }
  }
}
int main() {
  if (fopen("1213G"
            ".INP",
            "r")) {
    freopen(
        "1213G"
        ".INP",
        "r", stdin);
    freopen(
        "1213G"
        ".OUT",
        "w", stdout);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  scanf("%d %d", &n, &m);
  for (int i = 1, _n = (n); i <= _n; ++i) parent[i] = i, Rank[i] = 1;
  edge.resize(n - 1);
  int u, v, w, q;
  for (int i = 0, _n = (n - 1); i < _n; ++i)
    scanf("%d %d %d", &edge[i].second.first, &edge[i].second.second,
          &edge[i].first);
  sort(edge.begin(), edge.end());
  vector<query> qu(m);
  for (int i = 0, _n = (m); i < _n; ++i) {
    scanf("%d", &qu[i].Q);
    qu[i].id = i;
  }
  sort(qu.begin(), qu.end(), cmpQ);
  int j = 0;
  long long ans = 0;
  for (int i = 0, _n = (m); i < _n; ++i) {
    for (; j < n - 1; ++j) {
      if (edge[j].first <= qu[i].Q) {
        ans += (long long)Rank[Find(edge[j].second.first)] *
               Rank[Find(edge[j].second.second)];
        Union(edge[j].second.first, edge[j].second.second);
      } else
        break;
    }
    qu[i].res = ans;
  }
  sort(qu.begin(), qu.end(), ret);
  for (int i = 0, _n = (m); i < _n; ++i) cout << qu[i].res << ' ';
  return 0;
}