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;
long long int prnt[200005], sz[200005];
struct edge {
  long long int u, v, w;
};
bool comp(edge a, edge b) { return a.w < b.w; }
long long int root(long long int x) {
  while (x != prnt[x]) {
    prnt[x] = prnt[prnt[x]];
    x = prnt[x];
  }
  return x;
}
void merges(long long int x, long long int y) {
  if (sz[x] > sz[y]) {
    sz[x] += sz[y];
    prnt[y] = x;
  } else {
    sz[y] += sz[x];
    prnt[x] = y;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, i, j, w, u, v;
  cin >> n >> m;
  vector<edge> e(n - 1);
  vector<long long int> answer;
  long long int weight[200005] = {0};
  for (i = 0; i < n - 1; i++) {
    cin >> e[i].u >> e[i].v >> e[i].w;
  }
  sort(e.begin(), e.end(), comp);
  for (i = 0; i < 200005; i++) {
    prnt[i] = i;
    sz[i] = 1;
  }
  for (i = 0; i < n - 1; i++) {
    u = root(e[i].u);
    v = root(e[i].v);
    w = (e[i].w);
    if (u != v) {
      weight[w] += sz[u] * sz[v];
      merges(u, v);
    }
  }
  for (i = 1; i < 200005; i++) weight[i] += weight[i - 1];
  for (i = 0; i < m; i++) {
    cin >> w;
    answer.push_back(weight[w]);
  }
  for (i = 0; i < m; i++) cout << answer[i] << " ";
  cout << "\n";
  return 0;
}