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 p = 0;
int find(int parent[], int u) {
  if (parent[u] == -1) return u;
  parent[u] = find(parent, parent[u]);
  return parent[u];
}
void connect(int parent[], long long size[], int u, int v) {
  int a = find(parent, u);
  int b = find(parent, v);
  if (size[a] < size[b]) {
    p += size[a] * size[b];
    parent[a] = b;
    size[b] += size[a];
  } else {
    p += size[a] * size[b];
    parent[b] = a;
    size[a] += size[b];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  int mq = 0;
  tuple<int, int, int> g[n - 1];
  int parent[n];
  long long size[n];
  int q[m];
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--, w--;
    get<0>(g[i]) = w;
    get<1>(g[i]) = u;
    get<2>(g[i]) = v;
  }
  for (int i = 0; i < n; i++) {
    parent[i] = -1;
    size[i] = 1;
  }
  for (int i = 0; i < m; i++) {
    cin >> q[i];
    q[i]--;
  }
  for (int i = 0; i < m; i++) mq = max(mq, q[i]);
  long long ans[mq + 1];
  for (int i = 0; i <= mq; i++) ans[i] = 0;
  sort(g, g + n - 1);
  int j = 0;
  for (int i = 0; i <= mq; i++) {
    while (j < n && get<0>(g[j]) == i) {
      connect(parent, size, get<1>(g[j]), get<2>(g[j]));
      j++;
    }
    ans[i] = p;
  }
  for (int i = 0; i < m; i++) cout << ans[q[i]] << " ";
  cout << "\n";
}