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>
#pragma GCC optimize "03"
using namespace std;
const long long int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 1e9 + 9;
pair<pair<long long int, long long int>, long long int> e[N];
pair<long long int, long long int> q[N];
long long int ans, res[N], par[N], sz[N];
bool cmp(pair<pair<long long int, long long int>, long long int> a,
         pair<pair<long long int, long long int>, long long int> b) {
  return a.second < b.second;
}
long long int root(long long int x) {
  while (par[x] != x) x = par[par[x]];
  return x;
}
void merge(long long int x, long long int y) {
  x = root(x);
  y = root(y);
  ans -= sz[x] * (sz[x] - 1) / 2;
  ans -= sz[y] * (sz[y] - 1) / 2;
  ans += (sz[x] + sz[y]) * (sz[x] + sz[y] - 1) / 2;
  if (sz[x] < sz[y])
    par[x] = y, sz[y] += sz[x];
  else
    par[y] = x, sz[x] += sz[y];
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  for (long long int i = 1; i <= n - 1; i++)
    cin >> e[i].first.first >> e[i].first.second >> e[i].second;
  sort(e + 1, e + n, cmp);
  for (long long int i = 1; i <= m; i++) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(q + 1, q + m + 1);
  long long int j = 1;
  for (long long int i = 1; i <= m; i++) {
    while (j <= n - 1 && e[j].second <= q[i].first) {
      merge(e[j].first.first, e[j].first.second);
      j++;
    }
    res[q[i].second] = ans;
  }
  for (long long int i = 1; i <= m; i++) cout << res[i] << " ";
  return 0;
}