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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int64_t N = 228;
vector<int64_t> g[N];
int64_t c[N], dp[N][N];
int64_t n, k;
void dfs(int64_t v, int64_t p) {
  if (g[v].size() == 1 && g[v][0] == p) {
    dp[v][0] = c[v];
    return;
  }
  for (int64_t x : g[v])
    if (x != p) dfs(x, v);
  dp[v][0] = c[v];
  for (int64_t x : g[v]) {
    if (x != p) {
      int64_t mx = 0;
      for (int64_t kk = k - 1; kk <= 200; kk++) mx = max(mx, dp[x][kk]);
      dp[v][0] += mx;
    }
  }
  for (int64_t x : g[v]) {
    if (x == p) continue;
    for (int64_t kk = 0; kk <= k; kk++) {
      int64_t sum = dp[x][kk];
      int64_t f = k - kk - 2;
      if (2 * f + 2 < k) f = kk;
      for (int64_t j : g[v]) {
        if (j == x || j == p) continue;
        int64_t mx = 0;
        for (int64_t i = f; i <= 200; i++) mx = max(mx, dp[j][i]);
        sum += mx;
      }
      dp[v][kk + 1] = max(dp[v][kk + 1], sum);
    }
  }
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  k++;
  for (int64_t i = 0; i < n; i++) cin >> c[i];
  for (int64_t i = 0; i < n - 1; i++) {
    int64_t u, v;
    cin >> u >> v;
    u--;
    v--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0, 0);
  int64_t res = 0;
  for (int64_t i = 0; i < N; i++)
    for (int64_t j = 0; j < N; j++) res = max(res, dp[i][j]);
  cout << res;
}