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>
std::string infile = "input.txt";
std::string outfile = "output.txt";
using namespace std;
using ull = unsigned long long;
using ll = long long;
template <typename T, typename U>
using p = pair<T, U>;
template <typename T>
using v = vector<T>;
ll combination2(ll x) { return x * (x - 1) * 0.5; }
v<ll> dsu_size;
v<ll> dsu_parent;
void create_dsu(ll size) {
  dsu_parent.assign(size, 0);
  iota(dsu_parent.begin(), dsu_parent.end(), 0);
  dsu_size.assign(size, 1);
}
void make_set(ll v) {
  dsu_parent[v] = v;
  dsu_size[v] = 1ll;
}
ll find_set(ll v) {
  if (v == dsu_parent[v]) return v;
  return dsu_parent[v] = find_set(dsu_parent[v]);
}
void union_set(ll a, ll b, ll& res) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    res = res - combination2(dsu_size[a]) - combination2(dsu_size[b]) +
          combination2(dsu_size[a] + dsu_size[b]);
    if (dsu_size[a] < dsu_size[b]) {
      swap(a, b);
    }
    dsu_parent[b] = a;
    dsu_size[a] += dsu_size[b];
  }
}
struct edge {
  ll u, v, w;
  bool operator<(const edge& a) const { return w < a.w; }
};
int main() {
  ll n, m;
  cin >> n >> m;
  create_dsu(n);
  v<edge> tree(n - 1);
  for (ull i = 0; i < (n - 1); ++i) {
    cin >> tree[i].u >> tree[i].v >> tree[i].w;
    --tree[i].u;
    --tree[i].v;
  }
  v<p<ll, ll>> q(m);
  for (ull i = 0; i < (m); ++i) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  sort(tree.begin(), tree.end());
  v<ll> ans(m);
  ll res = 0;
  ull pos = 0;
  for (auto query : q) {
    while (pos < tree.size() && tree[pos].w <= query.first) {
      union_set(tree[pos].u, tree[pos].v, res);
      ++pos;
    }
    ans[query.second] = res;
  }
  for (auto a : ans) {
    cout << a << " ";
  }
  return 0;
}