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;
struct Disjoint {
  vector<int> f, sz;
  void init(int n) {
    f.resize(n + 1);
    sz.resize(n + 1);
    for (int i = 1; i <= n; i++) f[i] = i, sz[i] = 1;
  }
  int ft(int x) { return (f[x] == x) ? x : f[x] = ft(f[x]); }
  void unite(int x, int y) {
    int fx = ft(x);
    int fy = ft(y);
    f[fx] = fy;
    sz[fy] += sz[fx];
    sz[fx] = 0;
  }
  bool conj(int x, int y) { return ft(x) == ft(y); }
};
vector<pair<int, int> > gr[1 + N];
long long ans[1 + N];
long long g(int x) { return 1ll * x * (x - 1) / 2; }
void solve() {
  int n, m;
  cin >> n >> m;
  Disjoint comp;
  comp.init(n);
  for (int i = 1; i < n; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    gr[w].push_back({x, y});
  }
  for (int w = 1; w <= N; w++) {
    ans[w] = ans[w - 1];
    for (auto edge : gr[w]) {
      int x = edge.first;
      int y = edge.second;
      int tx = comp.ft(x);
      int ty = comp.ft(y);
      if (tx == ty) continue;
      ans[w] -= g(comp.sz[tx]);
      ans[w] -= g(comp.sz[ty]);
      comp.unite(tx, ty);
      ans[w] += g(comp.sz[ty]);
    }
  }
  while (m--) {
    int q;
    cin >> q;
    cout << ans[q] << " ";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) solve();
  return 0;
}