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;
template <class T>
inline bool setmin(T &a, T b) {
  if (a > b) return a = b, 1;
  return 0;
}
template <class T>
inline bool setmax(T &a, T b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <class T>
inline T fast(T a, T b, T mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
template <class T>
inline T russian(T a, T b, T mod) {
  long long res = 0;
  while (b) {
    if (b & 1) res = (res + a) % mod;
    a = (a + a) % mod;
    b >>= 1;
  }
  return res;
}
template <class T>
istream &operator>>(istream &os, vector<T> &container) {
  for (auto &u : container) os >> u;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &container) {
  for (auto &u : container) os << u << " ";
  return os;
}
template <typename T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
const long long T = 1 << 17;
const long long INF = 1e9 + 7;
const long long mod = 998244353;
const long long BIG_INF = 1e18 + 7;
const long long N = 2e5 + 7;
long long n, m;
template <int SIZE>
struct dsu {
  int rep[SIZE];
  long long roz[SIZE];
  dsu() {
    for (int i = 0; i < SIZE; i++) {
      rep[i] = i;
      roz[i] = 1;
    }
  }
  int Find(int x) {
    if (rep[x] != x) rep[x] = Find(rep[x]);
    return rep[x];
  }
  void Union(int x, int y) {
    x = Find(x), y = Find(y);
    roz[x] += roz[y];
    rep[y] = x;
  }
};
long long odp[N];
long long ans;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  dsu<N> D;
  vector<vector<int> > kraw;
  for (int i = 1; i < n; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    kraw.push_back({c, a, b});
  }
  sort((kraw).begin(), (kraw).end());
  int id = 0;
  for (int i = 0; i < N; i++) {
    while (id < (int)(kraw).size() && kraw[id][0] == i) {
      int a = kraw[id][1], b = kraw[id][2];
      a = D.Find(a);
      b = D.Find(b);
      ans += D.roz[a] * D.roz[b];
      D.Union(a, b);
      id++;
    }
    odp[i] = ans;
  }
  for (int i = 0; i < m; i++) {
    int a;
    cin >> a;
    cout << odp[a] << '\n';
  }
}