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;
int n, k, a[5010], dep[5010], dp[5010][5010], val[5010][5010];
vector<int> v[5010];
inline int getint() {
  int r = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    r = r * 10 + ch - '0';
    ch = getchar();
  }
  return r;
}
inline void dfs(int x, int pr) {
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != pr) {
      dfs(v[x][i], x);
      dep[x] = max(dep[x], dep[v[x][i]] + 1);
    }
  dp[x][0] = a[x];
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != pr) {
      val[v[x][i]][n] = dp[v[x][i]][n];
      for (int j = n - 1; j >= 0; j--)
        val[v[x][i]][j] = max(val[v[x][i]][j + 1], dp[v[x][i]][j]);
      dp[x][0] += val[v[x][i]][k - 1];
    }
  for (int i = 1; i <= dep[x]; i++) {
    int res = 0, id = max(i - 1, max(0, k - i - 1));
    for (int j = 0; j < v[x].size(); j++)
      if (v[x][j] != pr) {
        res += val[v[x][j]][id];
      }
    for (int j = 0; j < v[x].size(); j++)
      if (v[x][j] != pr) {
        dp[x][i] = max(dp[x][i], dp[v[x][j]][i - 1] + res - val[v[x][j]][id]);
      }
  }
}
signed main() {
  n = getint();
  k = getint();
  k++;
  for (int i = 1; i <= n; i++) a[i] = getint();
  for (int i = 1; i < n; i++) {
    int x, y;
    x = getint();
    y = getint();
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, -1);
  int ans = 0;
  for (int i = 0; i <= n; i++) ans = max(ans, dp[1][i]);
  printf("%d\n", ans);
  return 0;
}