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 root[200005];
long long siz[200005], ans[200005] = {0};
long long sum = 0;
void init(int n) {
  for (int i = 1; i <= n; i++) {
    root[i] = i;
    siz[i] = 1;
  }
}
int fin(int x) { return root[x] == x ? x : (root[x] = fin(root[x])); }
void Union(int a, int b) {
  sum = sum + (long long)siz[fin(a)] * (long long)siz[fin(b)];
  if (siz[fin(a)] < siz[fin(b)]) swap(a, b);
  siz[fin(a)] = siz[fin(a)] + siz[fin(b)];
  root[fin(b)] = fin(a);
}
vector<pair<int, int> > e[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int m, n;
  cin >> n >> m;
  init(n);
  for (int i = 0; i < n - 1; i++) {
    int c, d, v;
    cin >> c >> d >> v;
    e[v].push_back(make_pair(c, d));
  }
  for (int i = 1; i <= 200000; i++) {
    for (int j = 0; j < e[i].size(); j++) {
      int c = e[i][j].first, d = e[i][j].second;
      Union(c, d);
    }
    ans[i] = sum;
  }
  for (int i = 0; i < m; i++) {
    int u;
    cin >> u;
    cout << ans[u] << ' ';
  }
}