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 ans = 0;
long long p[200001], c[200001], sz[200001];
void pre() {
  for (int i = 0; i < 200001; i++) {
    p[i] = i;
    c[i] = 1;
    sz[i] = 1;
  }
}
int root(int val) {
  while (p[val] != val) {
    p[val] = p[p[val]];
    val = p[val];
  }
  return val;
}
void connect(int x, int y) {
  x = root(x);
  y = root(y);
  if (x != y) {
    if (sz[x] < sz[y]) {
      p[y] = x;
      sz[x] = sz[y] + 1;
    } else {
      p[x] = y;
      sz[y] = sz[x] + 1;
    }
    ans += c[x] * c[y];
    long long co = c[x];
    c[x] += c[y];
    c[y] += co;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pre();
  int n, m;
  cin >> n >> m;
  vector<pair<int, pair<int, int> > > s;
  for (int x, y, w, i = 1; i < n; i++) {
    cin >> x >> y >> w;
    s.push_back({w, {x, y}});
  }
  sort(s.begin(), s.end());
  vector<pair<int, int> > v;
  for (int x, i = 0; i < m; i++) {
    cin >> x;
    v.push_back({x, i});
  }
  int cur = 0;
  sort(v.begin(), v.end());
  long long a[m + 1];
  for (int i = 0; i < m; i++) {
    for (; cur < n - 1;) {
      if (s[cur].first <= v[i].first) {
        int x = s[cur].second.first;
        int y = s[cur].second.second;
        connect(x, y);
        cur++;
      } else
        break;
    }
    a[v[i].second] = ans;
  }
  for (int i = 0; i < m; i++) {
    cout << a[i] << " ";
  }
}