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 pr[200010];
long long sz[200010];
long long dp[200010];
long long ans[200010];
long long get(long long x) {
  if (x == pr[x]) {
    return x;
  }
  pr[x] = get(pr[x]);
  return pr[x];
}
void _union(long long x, long long y) {
  x = get(x);
  y = get(y);
  if (sz[x] == sz[y]) {
    sz[x]++;
  }
  if (sz[x] > sz[y]) {
    pr[y] = x;
    dp[x] += dp[y];
  } else {
    pr[x] = y;
    dp[y] += dp[x];
  }
}
vector<vector<long long>> v;
vector<pair<long long, long long>> quest;
bool cmp(vector<long long> a, vector<long long> b) { return a[2] < b[2]; }
map<long long, long long> m;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  long long x, y, z;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y >> z;
    pr[i] = i;
    sz[i] = 1;
    dp[i] = 1;
    v.push_back({x - 1, y - 1, z});
  }
  pr[n - 1] = n - 1;
  dp[n - 1] = 1;
  sz[n - 1] = 1;
  for (long long i = 0; i < q; i++) {
    cin >> x;
    quest.push_back({x, i});
  }
  sort(v.begin(), v.end(), cmp);
  sort(quest.begin(), quest.end());
  long long pans = 0;
  long long ind = 0;
  for (long long i = 0; i < q; i++) {
    while (ind < v.size() && v[ind][2] <= quest[i].first) {
      if (get(v[ind][0]) == get(v[ind][1])) {
        ind++;
        continue;
      }
      pans += (dp[get(v[ind][0])] * dp[get(v[ind][1])]);
      _union(v[ind][0], v[ind][1]);
      ind++;
    }
    ans[quest[i].second] = pans;
  }
  for (long long i = 0; i < q; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}