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;
const int MAX_N = 2e5 + 3;
long long n, m;
long long par[MAX_N], sz[MAX_N], nums[MAX_N];
vector<pair<int, pair<int, int>>> edges;
long long ans;
int rt(int u) { return u ^ par[u] ? par[u] = rt(par[u]) : u; }
void join(int u, int v) {
  if ((u = rt(u)) == (v = rt(v))) return;
  ans += sz[u] * sz[v];
  if (sz[u] < sz[v]) swap(u, v);
  par[v] = u, sz[u] += sz[v];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < (n); i++) par[i] = i, sz[i] = 1;
  for (int i = 0; i < (n - 1); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    edges.push_back({c, {a, b}});
  }
  sort(edges.begin(), edges.end());
  for (int i = 0; i < (n - 1); i++) {
    join(edges[i].second.first, edges[i].second.second);
    nums[edges[i].first] = ans;
  }
  for (int i = (1); i < (MAX_N); i++) nums[i] = max(nums[i], nums[i - 1]);
  for (int i = 0; i < (m); i++) {
    int x;
    cin >> x;
    cout << nums[x] << ' ';
  }
  cout << '\n';
}