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;
using ll = long long;
using llu = unsigned long long;
ll mod = 1e9 + 7;
ll mode = 998244353;
ll ans1 = 0;
ll ans[200001];
ll a[200001];
ll w[200001];
vector<pair<ll, pair<ll, ll>>> v;
vector<pair<ll, ll>> u;
ll root(ll k) {
  while (a[k] != k) {
    a[k] = a[a[k]];
    k = a[k];
  }
  return k;
}
void unions(ll x, ll y) {
  ll k = root(x);
  ll k1 = root(y);
  if (k == k1) return;
  ans1 -= (w[k1] * (w[k1] - 1)) / 2;
  ans1 -= (w[k] * (w[k] - 1)) / 2;
  if (w[k] > w[k1]) {
    w[k] += w[k1];
    ans1 += (w[k] * (w[k] - 1)) / 2;
    a[k1] = a[k];
  } else {
    w[k1] += w[k];
    ans1 += (w[k1] * (w[k1] - 1)) / 2;
    a[k] = a[k1];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t, n, m, i, j, k;
  for (i = 0; i <= 200000; i++) {
    a[i] = i;
    w[i] = 1;
  }
  cin >> n >> m;
  for (i = 0; i < n - 1; i++) {
    ll x, y, z;
    cin >> x >> y >> z;
    v.push_back({z, {x, y}});
  }
  for (i = 0; i < m; i++) {
    cin >> k;
    u.push_back({k, i});
  }
  sort(u.begin(), u.end());
  sort(v.begin(), v.end());
  i = 0;
  for (j = 0; j < m; j++) {
    while (i < n - 1 && u[j].first >= v[i].first) {
      unions(v[i].second.first, v[i].second.second);
      i++;
    }
    ans[u[j].second] = ans1;
  }
  for (i = 0; i < m; i++) cout << ans[i] << " ";
}