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 fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long find(long long parent[], long long i) {
  if (parent[i] == -1) return i;
  return find(parent, parent[i]);
}
void Union(long long parent[], long long x, long long y) {
  long long xset = find(parent, x);
  long long yset = find(parent, y);
  if (xset != yset) {
    parent[xset] = yset;
  }
}
int main() {
  long long n, m, a, c, b;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  pair<long long, pair<long long, long long> > p[n - 1];
  for (long long i = (0); i < (n - 1); i++) {
    cin >> a >> b >> c;
    p[i] = make_pair(c, make_pair(a, b));
  }
  pair<long long, long long> arr[m];
  for (long long i = (0); i < (m); i++) {
    cin >> arr[i].first;
    arr[i].second = i;
  }
  long long ans[m], parent[n + 1];
  for (long long i = (1); i < (n + 1); i++) parent[i] = -1;
  sort(arr, arr + m);
  sort(p, p + n - 1);
  int j = 0;
  long long res = 0;
  long long visit[n + 1];
  for (long long i = (1); i < (1 + n); i++) visit[i] = 1;
  for (long long i = (0); i < (m); i++) {
    a = arr[i].first;
    b = arr[i].second;
    while (p[j].first <= a and j < n) {
      long long x = find(parent, p[j].second.first);
      long long y = find(parent, p[j].second.second);
      if (visit[x] > visit[y]) swap(x, y);
      res = res - (visit[x] * (visit[x] - 1)) / 2 -
            (visit[y] * (visit[y] - 1)) / 2 +
            ((visit[x] + visit[y]) * (visit[x] + visit[y] - 1)) / 2;
      visit[y] = visit[x] + visit[y];
      visit[x] = 0;
      parent[x] = y;
      j++;
    }
    ans[b] = res;
  }
  for (long long i = (0); i < (m); i++) cout << ans[i] << " ";
  return 0;
}