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 long long mod = 1e9 + 7;
const int N = 2e5 + 4;
int parent[N], rnk[N], siz[N];
long long ans;
int find(int node) {
  parent[node] = (parent[node] == node) ? node : find(parent[node]);
  return parent[node];
}
long long calc(int n) {
  long long val = (long long)n;
  val *= (val - 1);
  val /= 2;
  return val;
}
void merge(int i, int j) {
  int pi = find(i), pj = find(j);
  ans -= (calc(siz[pj]) + calc(siz[pi]));
  if (rnk[pi] > pj) {
    parent[pj] = pi;
    siz[pi] += siz[pj];
    ans += calc(siz[pi]);
  } else {
    parent[pi] = pj;
    if (rnk[pi] == rnk[pj]) rnk[pj]++;
    siz[pj] += siz[pi];
    ans += calc(siz[pj]);
  }
}
bool comp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.second < b.second;
}
void init() {
  for (int i = 1; i < N; i++) {
    parent[i] = i;
    rnk[i] = 0;
    siz[i] = 1;
  }
}
int main() {
  init();
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  vector<pair<pair<int, int>, int>> v;
  for (int i = 0; i < n - 1; i++) {
    int a, b, w;
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &w);
    v.push_back(make_pair(make_pair(a, b), w));
  }
  long long final[N], ind = 0;
  sort(v.begin(), v.end(), comp);
  for (int i = 1; i < N; i++) {
    for (; ind < v.size() && v[ind].second <= i; ind++) {
      int x = (v[ind].first).first, y = (v[ind].first).second;
      merge(x, y);
    }
    final[i] = ans;
  }
  while (m--) {
    int wt;
    scanf("%d", &wt);
    printf("%lld ", final[wt]);
  }
  return 0;
}