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;
vector<long long> par, cnt;
int find(int x) {
  if (x == par[x]) return par[x];
  par[x] = find(par[x]);
  return par[x];
}
long long get(long long val) { return (val * (val - 1)) / 2; }
void merge(long long a, long long b) {
  int para = find(a);
  int parb = find(b);
  if (para != parb) {
    ans = ans - get(cnt[para]) - get(cnt[parb]);
    par[para] = parb;
    cnt[parb] += cnt[para];
    cnt[para] = 0;
    ans += get(cnt[parb]);
  }
}
inline void solve() {
  int n, m;
  cin >> n >> m;
  par.resize(n + 1);
  cnt.resize(n + 1);
  vector<pair<long long, pair<long long, long long>>> edge;
  for (int i = 0, a, b, c; i < n - 1; i++) {
    cin >> a >> b >> c;
    edge.push_back({c, {a, b}});
  }
  sort(edge.begin(), edge.end());
  vector<long long> q(m);
  vector<long long> pos;
  pos.resize(m);
  for (int i = 0; i < m; i++) {
    cin >> q[i];
    pos[q[i] - 1] = i;
  }
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    cnt[i] = 1;
  }
  vector<long long> v(m + 1, 0);
  int mx = INT_MIN;
  for (int i = 0; i < edge.size(); i++) {
    int cst = edge[i].first, a = edge[i].second.first,
        b = edge[i].second.second;
    int para = find(a);
    int parb = find(b);
    mx = max(mx, cst);
    if (para != parb) {
      merge(para, parb);
    }
    v[mx] = ans;
  }
  for (int i = 1; i <= m; i++) {
    if (v[i] == 0) v[i] = v[i - 1];
  }
  for (int i = 0; i < m; i++) cout << v[q[i]] << " ";
}
int main() {
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}