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 N = 2e5 + 5, mod = 1e18 + 9;
int n, m;
pair<pair<int, int>, int> a[N];
int q[N];
long long dp[N];
int dad[N], ran[N], tam[N];
long long res;
vector<pair<int, int>> g[N];
int bus(int u) {
  if (dad[u] != u) dad[u] = bus(dad[u]);
  return dad[u];
}
void unir(int u, int v) {
  u = bus(u);
  v = bus(v);
  if (ran[u] > ran[v]) {
    dad[v] = u;
    tam[u] += tam[v];
  } else if (ran[v] > ran[u]) {
    dad[u] = v;
    tam[v] += tam[u];
  } else {
    dad[u] = v;
    tam[v] += tam[u];
    ran[v]++;
  }
  return;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    dad[i] = i;
    ran[i] = 0;
    tam[i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> a[i].first.first >> a[i].first.second >> a[i].second;
    g[a[i].second].push_back({a[i].first});
  }
  for (int i = 1; i <= m; i++) cin >> q[i];
  for (int i = 1; i <= 200000; i++) {
    for (auto u : g[i]) {
      if (bus(u.first) != bus(u.second)) {
        long long p1, p2;
        p1 = tam[bus(u.first)];
        p2 = tam[bus(u.second)];
        p1 *= (p1 - 1);
        p1 /= 2;
        p2 *= (p2 - 1);
        p2 /= 2;
        res -= p1;
        res -= p2;
        unir(u.first, u.second);
        p1 = tam[bus(u.first)];
        p1 *= (p1 - 1);
        p1 /= 2;
        res += p1;
      }
    }
    dp[i] = res;
  }
  for (int i = 1; i <= m; i++) cout << dp[q[i]] << " ";
  return 0;
}