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;
  edge() {}
  edge(int i, int j, int ww) { u = i, j = v, w = ww; }
};
long long int sz[200005];
int par[200005];
vector<long long int> res(200005);
long long int ans = 0;
int Root(int a) {
  while (a != par[a]) a = par[a];
  return a;
}
void Union(int a, int b) {
  int ra = Root(a);
  int rb = Root(b);
  if (ra == rb) return;
  ans += sz[ra] * sz[rb];
  if (sz[ra] > sz[rb]) {
    par[rb] = ra;
    sz[ra] += sz[rb];
  } else {
    par[ra] = rb;
    sz[rb] += sz[ra];
  }
}
bool comp(edge o1, edge o2) { return o1.w < o2.w; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  for (int i = 0; i < 200005; i += 1) par[i] = i, sz[i] = 1;
  int n, m;
  cin >> n >> m;
  vector<edge> a(n - 1);
  for (int i = 0; i < n - 1; i += 1) cin >> a[i].u >> a[i].v >> a[i].w;
  sort(a.begin(), a.end(), comp);
  vector<pair<int, int> > q(m);
  for (int i = 0; i < m; i += 1) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  int ptr = 0;
  for (int i = 0; i < m; i += 1) {
    int wt = q[i].first;
    int j = q[i].second;
    while (ptr < n - 1 && a[ptr].w <= wt) {
      Union(a[ptr].u, a[ptr].v);
      ptr++;
    }
    res[j] = ans;
  }
  for (int i = 0; i < m; i += 1) cout << res[i] << ' ';
  cout << '\n';
}