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;
const int N = 2e5 + 7;
const int Mod = 1e9 + 7;
const int INF = 2e9 + 7;
void in() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
struct haha {
  int u, v, w;
  bool operator<(const haha& x) const { return w < x.w; }
};
int par[N];
struct DSU {
  void init(int n) {
    for (int i = 1; i <= n; i++) par[i] = -1;
  }
  int root(int u) { return par[u] < 0 ? u : par[u] = root(par[u]); }
  void join(int u, int v) {
    u = root(u);
    v = root(v);
    if (u == v) return;
    if (par[u] > par[v]) swap(u, v);
    par[u] += par[v];
    par[v] = u;
  }
} dsu;
haha a[N];
int query[N];
int pos[N];
long long f[N];
int main() {
  in();
  int n, q;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    cin >> a[i].u >> a[i].v >> a[i].w;
  }
  for (int i = 1; i <= q; i++) {
    cin >> query[i];
    pos[i] = query[i];
  }
  sort(a + 1, a + n);
  sort(query + 1, query + 1 + q);
  dsu.init(n);
  int j = 1;
  long long res = 0;
  for (int i = 1; i <= q; i++) {
    while (j < n && a[j].w <= query[i]) {
      array<long long, 2> _size = {-par[dsu.root(a[j].u)],
                                   -par[dsu.root(a[j].v)]};
      res -= _size[1] * (_size[1] - 1) / 2 + _size[0] * (_size[0] - 1) / 2;
      dsu.join(a[j].u, a[j].v);
      int size = -par[dsu.root(a[j].u)];
      res += 1LL * size * (size - 1) / 2;
      j++;
    }
    f[query[i]] = res;
  }
  for (int i = 1; i <= q; i++) cout << f[pos[i]] << " ";
  return 0;
}