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;
struct edge {
  long long f, t, w;
  bool operator<(edge x) const { return w < x.w; }
} e[200005];
struct query {
  long long w, i, ans;
  bool operator<(query x) const { return w < x.w; }
} qq[200005];
long long p[200005], num[200005];
long long find(long long x) { return p[x] ? p[x] = find(p[x]) : x; }
bool cmp(query x, query y) { return x.i < y.i; }
int main() {
  long long n, q, i, j, k, t;
  cin >> n >> q;
  num[n] = 1;
  for (i = 1; i < n; i++) {
    cin >> e[i].f >> e[i].t >> e[i].w;
    num[i] = 1;
  }
  for (i = 1; i <= q; i++) {
    cin >> qq[i].w;
    qq[i].i = i;
  }
  sort(e + 1, e + n);
  sort(qq + 1, qq + i);
  for (i = j = 1; j <= q; j++) {
    qq[j].ans = qq[j - 1].ans;
    while (i < n && e[i].w <= qq[j].w) {
      k = find(e[i].f);
      t = find(e[i].t);
      qq[j].ans -= (num[t] * (num[t] - 1) >> 1) + (num[k] * (num[k] - 1) >> 1);
      num[k] += num[t];
      p[t] = k;
      qq[j].ans += num[k] * (num[k] - 1) >> 1;
      i++;
    }
  }
  sort(qq + 1, qq + q + 1, cmp);
  for (i = 1; i <= q; i++) cout << qq[i].ans << (i == q ? '\n' : ' ');
  return 0;
}