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 mod = 1e9 + 7;
const long long inf = 3e18 + 5;
int add(int a, int b) { return (a += b) < mod ? a : a - mod; }
int mul(int a, int b) { return 1LL * a * b % mod; }
struct UF {
  vector<int> p;
  vector<long long> sz;
  int comps;
  UF(int s) : comps(s) {
    sz.resize(comps + 1);
    p.resize(comps + 1);
    init();
  }
  void init() {
    for (int i = 0; i <= comps; i++) {
      sz[i] = 1;
      p[i] = i;
    }
  }
  int fset(int u) { return u == p[u] ? u : p[u] = fset(p[u]); }
  bool connect(int u, int v, long long& x) {
    u = fset(u);
    v = fset(v);
    if (u != v) {
      if (sz[v] > sz[u]) {
        swap(u, v);
      }
      x -= sz[v] * (sz[v] + 1) / 2;
      x -= sz[u] * (sz[u] + 1) / 2;
      sz[u] += sz[v];
      x += sz[u] * (sz[u] + 1) / 2;
      p[v] = u;
      comps--;
      return true;
    }
    return false;
  }
};
long long sol[200600];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  UF uf(n);
  vector<tuple<int, int, int>> a;
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    a.emplace_back(w, u, v);
  }
  sort(a.begin(), a.end());
  vector<int> query(q);
  for (int i = 0; i < q; i++) {
    cin >> query[i];
  }
  vector<int> qq = query;
  sort(qq.begin(), qq.end());
  int i = 0;
  long long res = 0;
  for (int x : qq) {
    for (int j = i; j < n - 1; j++) {
      auto [w, u, v] = a[j];
      if (w > x) {
        i = j;
        break;
      }
      uf.connect(u, v, res);
      if (j == n - 2) i = j + 1;
    }
    sol[x] = res;
  }
  for (int x : query) {
    cout << sol[x] << " ";
  }
}