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;
struct edge {
  int u, v, w;
};
bool cmp(const edge& a, const edge& b) { return a.w > b.w; }
int find(int parent[], int i) {
  if (parent[i] == i)
    return i;
  else
    return parent[i] = find(parent, parent[i]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<edge> edges(n - 1);
  for (int i = 0; i < n - 1; i++) {
    cin >> edges[i].u >> edges[i].v >> edges[i].w;
    edges[i].u--;
    edges[i].v--;
  }
  vector<pair<int, int>> queries(m);
  for (int i = 0; i < m; i++) {
    cin >> queries[i].first;
    queries[i].second = i;
  }
  int parent[n], rank[n], size[n];
  for (int i = 0; i < n; i++) {
    parent[i] = i;
    rank[i] = 0;
    size[i] = 1;
  }
  sort(queries.begin(), queries.end());
  sort(edges.begin(), edges.end(), cmp);
  long long res = 0;
  vector<long long> ans(m);
  for (int i = 0; i < m; i++) {
    while (!edges.empty() && edges.back().w <= queries[i].first) {
      auto e = edges.back();
      edges.pop_back();
      int u = e.u;
      int v = e.v;
      int w = e.w;
      int p1 = find(parent, u);
      int p2 = find(parent, v);
      if (p1 == p2) continue;
      res = res - size[p1] * ((long long)size[p1] - 1) / 2 -
            size[p2] * ((long long)size[p2] - 1) / 2 +
            ((long long)size[p1] + size[p2]) *
                ((long long)size[p1] + size[p2] - 1) / 2;
      if (rank[p1] > rank[p2]) {
        parent[p2] = p1;
        size[p1] += size[p2];
      } else {
        parent[p1] = p2;
        size[p2] += size[p1];
        if (rank[p1] == rank[p2]) rank[p2]++;
      }
    }
    ans[queries[i].second] = res;
  }
  for (auto v : ans) cout << v << ' ';
}