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;
int n, m;
int p[200005];
int sz[200005];
long long ans[200005];
pair<pair<int, int>, int> e[200005];
inline bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return x.second < y.second;
}
int get(int x) {
  if (x == p[x]) return x;
  return p[x] = get(p[x]);
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i < n; ++i) {
    cin >> e[i].first.first >> e[i].first.second >> e[i].second;
  }
  sort(e + 1, e + n, cmp);
  for (int i = 1; i <= n; ++i) p[i] = i, sz[i] = 1;
  for (int i = 1; i < n; ++i) {
    int x, y;
    x = e[i].first.first;
    y = e[i].first.second;
    x = get(x);
    y = get(y);
    ans[i] = ans[i - 1] + sz[x] * 1ll * sz[y];
    p[y] = x;
    sz[x] += sz[y];
  }
  while (m--) {
    int x;
    cin >> x;
    int l = 0;
    int r = n - 1;
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (e[mid].second <= x)
        l = mid;
      else
        r = mid - 1;
    }
    cout << ans[l] << " ";
  }
  cout << '\n';
  return 0;
}