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 UFelement {
 public:
  int rodic, rank;
  UFelement() {}
  UFelement(int rodic, int rank) {
    this->rodic = rodic;
    this->rank = rank;
  }
};
vector<UFelement> UF;
void createUF(int size) { UF.assign(size, UFelement(-1, 1)); }
void unionUtil(int a, int b) {
  if (UF[a].rank > UF[b].rank) {
    UF[b].rodic = a;
    UF[a].rank += UF[b].rank;
  } else {
    UF[a].rodic = b;
    UF[b].rank += UF[a].rank;
  }
}
int findd(int i) {
  int p = UF[i].rodic;
  if (p == -1) return i;
  int odp = findd(p);
  if (odp == -1) return p;
  UF[i].rodic = odp;
  return odp;
}
void unionn(int a, int b) { unionUtil(findd(a), findd(b)); }
struct Hrana {
 public:
  int a, b, cena;
  Hrana() {}
  Hrana(int a, int b, int cena) {
    this->a = a;
    this->b = b;
    this->cena = cena;
  }
  bool operator<(const Hrana& str) const { return (cena < str.cena); }
};
int T, Q, N, M, a, b;
vector<Hrana> hrany;
vector<int> input, og;
int main(int argc, char** argv) {
  cin >> N >> M;
  for (int i = 1; i < N; i++) {
    cin >> a >> b >> T;
    hrany.push_back(Hrana(a - 1, b - 1, T));
  }
  for (int i = 0; i < M; i++) {
    cin >> a;
    og.push_back(a);
    input.push_back(a);
  }
  sort(hrany.begin(), hrany.end());
  sort(input.begin(), input.end());
  createUF(N);
  map<int, unsigned long long int> odp;
  unsigned long long int res = 0;
  int odpI = 0;
  for (int i = 0; i < N - 1 && hrany[i].cena <= input[M - 1]; i++) {
    while (hrany[i].cena > input[odpI]) {
      odp[input[odpI]] = res;
      odpI++;
    }
    int fA = findd(hrany[i].a);
    int fB = findd(hrany[i].b);
    if (fA == fB) continue;
    res += UF[fA].rank * 1ULL * UF[fB].rank;
    unionn(fA, fB);
  }
  for (; odpI < M; odpI++) odp[input[odpI]] = res;
  cout << odp[og[0]];
  for (int i = 1; i < M; i++) cout << " " << odp[og[i]];
  cout << endl;
  return 0;
}