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;
using LL = long long;
namespace _buff {
const size_t BUFF = 1 << 19;
char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
char getc() {
  if (ib == ie) {
    ib = ibuf;
    ie = ibuf + fread(ibuf, 1, BUFF, stdin);
  }
  return ib == ie ? -1 : *ib++;
}
}  // namespace _buff
LL read() {
  using namespace _buff;
  LL ret = 0;
  bool pos = true;
  char c = getc();
  for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
    assert(~c);
  }
  if (c == '-') {
    pos = false;
    c = getc();
  }
  for (; c >= '0' && c <= '9'; c = getc()) {
    ret = (ret << 3) + (ret << 1) + (c ^ 48);
  }
  return pos ? ret : -ret;
}
int main() {
  int n = read(), m = read();
  vector<int> par(n + 1), siz(n + 1, 1);
  iota(par.begin(), par.end(), 0);
  function<int(int)> find = [&](int x) {
    return par[x] == x ? x : (x = find(par[x]));
  };
  function<LL(int, int)> merge = [&](int x, int y) {
    x = find(x);
    y = find(y);
    assert(x != y);
    LL ret = (LL)siz[x] * siz[y];
    if (siz[x] < siz[y]) swap(x, y);
    par[y] = x;
    siz[x] += siz[y];
    return ret;
  };
  vector<tuple<int, int, int> > vc;
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read(), w = read();
    vc.emplace_back(w, u, v);
  }
  sort(vc.begin(), vc.end());
  auto foo = vc.begin();
  vector<int> ask(m), idx(m);
  vector<LL> ans(m);
  iota(idx.begin(), idx.end(), 0);
  generate_n(ask.begin(), m, read);
  sort(idx.begin(), idx.end(), [&](int x, int y) { return ask[x] < ask[y]; });
  LL cur = 0;
  for (int id : idx) {
    int w = ask[id];
    for (; foo != vc.end() && get<0>(*foo) <= w; ++foo) {
      cur += merge(get<1>(*foo), get<2>(*foo));
    }
    ans[id] = cur;
  }
  for (LL x : ans) printf("%lld\n", x);
  return 0;
}