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 inf = 1e18;
const long long Mod = 1e9 + 7;
const long double Pi = acos(-1);
int n, k;
vector<int> v[209];
int a[209];
int dp[209][209];
void dfs(int node, int p) {
  dp[node][0] = a[node];
  for (auto u : v[node]) {
    if (u == p) continue;
    dfs(u, node);
    dp[node][0] += dp[u][k - 1];
  }
  for (int j = 1; j <= n; j++) {
    if (k - j - 1 < 0) break;
    for (auto u : v[node]) {
      if (u == p) continue;
      int crnt = dp[u][j - 1];
      for (auto x : v[node]) {
        if (x == p || x == u) continue;
        crnt += dp[x][max(j - 1, k - j - 1)];
      }
      dp[node][j] = max(dp[node][j], crnt);
    }
  }
  for (int i = n - 1; i >= 0; i--)
    dp[node][i] = max(dp[node][i], dp[node][i + 1]);
}
int main() {
  cin >> n >> k;
  k++;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(0, 0);
  int mx = 0;
  for (int i = 0; i <= n; i++) mx = max(mx, dp[0][i]);
  cout << mx << endl;
}