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 n, k, a[205];
long long ee, h[205], nex[205 << 1], to[205 << 1];
long long f[205][205];
void addedge(long long x, long long y) {
  nex[++ee] = h[x];
  to[ee] = y;
  h[x] = ee;
}
void dfs(long long x, long long pre) {
  f[x][0] = a[x];
  for (long long i = h[x]; i; i = nex[i])
    if (to[i] != pre) dfs(to[i], x);
  for (long long i = h[x]; i; i = nex[i])
    if (to[i] != pre) f[x][0] += f[to[i]][k - 1];
  for (long long dep = 1; dep < n; dep++) {
    for (long long i = h[x]; i; i = nex[i])
      if (to[i] != pre) {
        long long s = 0;
        for (long long j = h[x]; j; j = nex[j])
          if (to[j] != pre && i != j) s += f[to[j]][max(dep - 1, k - dep - 1)];
        f[x][dep] = max(f[x][dep], s + f[to[i]][dep - 1]);
      }
  }
  for (long long dep = n - 1; dep > 0; dep--)
    f[x][dep - 1] = max(f[x][dep - 1], f[x][dep]);
}
signed main() {
  cin >> n >> k;
  k++;
  for (long long i = 1; i <= n; scanf("%lld", a + i), i++)
    ;
  for (long long i = 1, x, y; i < n && scanf("%lld%lld", &x, &y); i++)
    addedge(x, y), addedge(y, x);
  dfs(1, 0);
  cout << f[1][0] << endl;
  return 0;
}