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 arr[200005], visited[200005], result = 0, res[200005], p = 0;
struct graph {
  long long u, v, wt;
};
struct query {
  long long u, ind;
};
query arr2[200005];
graph adj[200005];
bool comparator(graph u, graph v) { return u.wt < v.wt; }
bool comparator2(query i, query j) { return i.u < j.u; }
long long find(long long u) {
  if (visited[u] < 0) return u;
  return visited[u] = find(visited[u]);
}
void solve(long long n, long long m, long long x) {
  while (adj[p].wt <= n && p <= x) {
    long long a = find(adj[p].u);
    long long b = find(adj[p].v);
    long long i = abs(visited[a]), j = abs(visited[b]);
    if (i > j) {
      visited[a] = -(i + j);
      visited[b] = a;
    } else {
      visited[b] = -(i + j);
      visited[a] = b;
    }
    result = result - ((i) * (i - 1)) / 2 - ((j) * (j - 1)) / 2 +
             ((i + j) * (i + j - 1)) / 2;
    p++;
  }
  res[m] = result;
}
int main() {
  long long i, j, k, u, v, n, q, a, b, c;
  cin >> n >> q;
  for (i = 1; i <= n - 1; i++) cin >> adj[i].u >> adj[i].v >> adj[i].wt;
  sort(adj + 1, adj + n, comparator);
  memset(visited, -1, sizeof(visited));
  for (i = 1; i <= q; i++) {
    long long s;
    cin >> s;
    arr2[i].u = s;
    arr2[i].ind = i;
  }
  sort(arr2 + 1, arr2 + q + 1, comparator2);
  for (i = 1; i <= q; i++) solve(arr2[i].u, arr2[i].ind, n - 1);
  for (i = 1; i <= q; i++) cout << res[i] - 1 << " ";
  cout << endl;
}