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 int maxn = 2e3 + 5;
const int mod = 1e9 + 7;
int n, k, a[maxn], dp[maxn][maxn];
;
vector<int> e[maxn];
void dfs(int x, int p) {
  int s[202];
  for (int i = (0); i < (n + 1); i++) s[i] = 0;
  for (auto u : e[x]) {
    if (u == p) continue;
    dfs(u, x);
    for (int i = (0); i < (n + 1); i++) {
      s[i] += dp[u][i];
    }
  }
  for (auto u : e[x]) {
    if (u == p) continue;
    for (int i = (0); i < (n + 1); i++) {
      int p = k - i - 1;
      if (p < i) p = i;
      dp[x][i + 1] = max(dp[x][i + 1], dp[u][i] + s[p] - dp[u][p]);
    }
  }
  dp[x][0] = s[k] + a[x];
  for (int i = n; i >= 0; i--) dp[x][i] = max(dp[x][i + 1], dp[x][i]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (1); i < (n + 1); i++) cin >> a[i];
  for (int i = (1); i < (n); i++) {
    int x, y;
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  int ans = 0;
  for (int i = (1); i < (n + 1); i++) {
    memset(dp, 0, sizeof(dp));
    dfs(i, 0);
    ans = max(ans, dp[i][0]);
  }
  cout << ans;
  return 0;
}