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 sz[200005];
long long cac[200005];
vector<pair<long long, pair<long long, long long> > > edges;
vector<pair<long long, long long> > ask;
long long ans[200005];
long long cur = 0;
long long find(long long i) {
  return cac[i] == i ? cac[i] : cac[i] = find(cac[i]);
}
void axcs(long long u, long long v) {
  long long cu = find(u);
  long long cv = find(v);
  cur += sz[cv] * sz[cu];
  sz[cv] += sz[cu];
  cac[cu] = cv;
}
signed main() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cac[i] = i;
    sz[i] = 1;
  }
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v, val;
    cin >> u >> v >> val;
    edges.push_back({val, {u, v}});
  }
  sort(edges.begin(), edges.end());
  for (long long i = 1; i <= m; i++) {
    long long cacc;
    cin >> cacc;
    ask.push_back({cacc, i});
  }
  memset(ans, -1, sizeof(ans));
  sort(ask.begin(), ask.end());
  long long it = 0;
  for (long long i = 0; i < n - 1; i++) {
    while (it < ask.size() && ask[it].first < edges[i].first) {
      ans[ask[it].second] = cur;
      it++;
    }
    axcs(edges[i].second.first, edges[i].second.second);
  }
  for (long long i = 1; i <= m; i++) {
    if (ans[i] == -1) {
      cout << cur;
    } else {
      cout << ans[i];
    }
    cout << " ";
  }
}