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;
long long int n, m;
long long int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long int dy[] = {0, -1, -1, -1, 0, 1, 1, 1};
long long int dxx[] = {1, -1, 0, 0};
long long int dyy[] = {0, 0, -1, 1};
long long int modexpo(long long int x, long long int y) {
  if (y == 0) return 1;
  if (y % 2) {
    long long int viky = modexpo(x, y / 2);
    return (((x * viky) % 998244353) * viky) % 998244353;
  } else {
    long long int viky = modexpo(x, y / 2);
    return (viky * viky) % 998244353;
  }
}
bool isvalid(long long int x, long long int y) {
  return (x >= 1 && x <= n && y > 0 && y <= m);
}
vector<pair<int, int> > v[200005];
long long int a[200005];
long long int sz[200005];
long long int curr = 0;
void init() {
  for (int i = 1; i < 200005; i++) {
    a[i] = i;
    sz[i] = 1;
  }
}
int root(int x) {
  while (a[x] != x) {
    x = a[x];
    a[x] = a[a[x]];
  }
  return x;
}
void join(int x, int y) {
  int px = root(x);
  int py = root(y);
  curr += sz[px] * sz[py];
  if (sz[px] < sz[py]) {
    a[px] = py;
    sz[py] += sz[px];
  } else {
    a[py] = px;
    sz[px] += sz[py];
  }
}
long long int ans[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    v[w].push_back(make_pair(x, y));
  }
  curr = 0;
  ans[0] = curr;
  init();
  for (int i = 1; i <= 200000; i++) {
    if (v[i].size() == 0) {
      ans[i] = ans[i - 1];
      continue;
    }
    for (int j = 0; j < v[i].size(); j++) {
      int x = v[i][j].first;
      int y = v[i][j].second;
      join(x, y);
    }
    ans[i] = curr;
  }
  while (q--) {
    int x;
    cin >> x;
    cout << ans[x] << " ";
  }
  cout << endl;
  return 0;
}