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 dp[205][205], dp1[205], n, x, y, a, K;
vector<int> v[205];
void solve(int nd, int pd) {
  for (int i = 0; i < (int)v[nd].size(); ++i) {
    int newn = v[nd][i];
    if (newn == pd) continue;
    solve(newn, nd);
    fill(dp1, dp1 + n + 1, 0);
    for (int j = 0; j <= n; ++j)
      for (int k = 0; k <= n; ++k)
        if (j + k + 1 > K)
          dp1[min(k + 1, j)] = max(dp1[min(k + 1, j)], dp[nd][j] + dp[newn][k]);
    for (int j = 0; j <= n; ++j) dp[nd][j] = max(dp[nd][j], dp1[j]);
  }
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; ++i) scanf("%d", &dp[i][0]);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  solve(1, -1);
  int sol = -1;
  for (int i = 0; i <= n; ++i) sol = max(sol, dp[1][i]);
  printf("%d\n", sol);
  return 0;
}