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;
vector<int> parent, level;
int find(int v) {
  if (v == parent[v]) return v;
  return parent[v] = find(parent[v]);
}
long long res;
long long sum(int cnt) { return 1LL * cnt * (cnt - 1) / 2; }
void merge(int u, int v) {
  u = find(u);
  v = find(v);
  if (level[u] < level[v]) swap(u, v);
  res -= sum(level[u]);
  res -= sum(level[v]);
  level[u] += level[v];
  res += sum(level[u]);
  parent[v] = u;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  res = 0;
  parent = level = vector<int>(n, 1);
  iota(parent.begin(), parent.end(), 0);
  vector<pair<int, pair<int, int>>> e(n - 1);
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    u--;
    v--;
    e[i].first = w;
    e[i].second.first = u;
    e[i].second.second = v;
  }
  vector<pair<int, int>> q(m);
  vector<long long> ans(m);
  for (int i = 0; i < m; ++i) {
    int temp_q;
    scanf("%d", &temp_q);
    q[i].first = temp_q;
    q[i].second = i;
  }
  sort(e.begin(), e.end());
  sort(q.begin(), q.end());
  int pos = 0;
  for (int i = 0; i < m; ++i) {
    while (pos < n - 1 && e[pos].first <= q[i].first) {
      int u = e[pos].second.first;
      int v = e[pos].second.second;
      merge(u, v);
      ++pos;
    }
    ans[q[i].second] = res;
  }
  for (int i = 0; i < m; ++i) {
    printf("%lld ", ans[i]);
  }
}