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 = 205;
int n, k, depest;
int wei[maxn], dp[maxn][maxn];
vector<int> v[maxn];
void findepth(int x, int dep, int par) {
  depest = max(dep, depest);
  for (int i = 0; i <= (int)v[x].size() - 1; i++)
    if (v[x][i] != par) findepth(v[x][i], dep + 1, x);
}
void dfs(int x, int par) {
  for (int i = 0; i <= (int)v[x].size() - 1; i++)
    if (v[x][i] != par) dfs(v[x][i], x);
  for (int dep = 0; dep <= depest; dep++)
    if (!dep) {
      for (int i = 0; i <= (int)v[x].size() - 1; i++)
        if (v[x][i] != par) dp[x][dep] += dp[v[x][i]][k - 1];
    } else {
      for (int i = 0; i <= (int)v[x].size() - 1; i++) {
        int y = v[x][i], z, now = dp[y][dep - 1];
        if (y == par) continue;
        for (int j = 0; j <= (int)v[x].size() - 1; j++) {
          z = v[x][j];
          if (z != par && z != y) now += dp[z][max(dep - 1, k - dep - 1)];
        }
        dp[x][dep] = max(dp[x][dep], now);
      }
    }
  for (int dep = depest; dep >= 1; dep--)
    dp[x][dep - 1] = max(dp[x][dep - 1], dp[x][dep]);
}
int main() {
  cin >> n >> k;
  k++;
  for (int i = 1; i <= n; i++) {
    cin >> wei[i];
    dp[i][0] = wei[i];
  }
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  findepth(1, 1, 0);
  dfs(1, 0);
  cout << dp[1][0];
  return 0;
}