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 N = 222;
const int INF = 0x3f3f3f3f;
int n, k, ans = 0;
int a[N];
int dp[N][N], f[N][N];
vector<int> adj[N];
void Read_Input() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
void DFS(int u, int p, int depth) {
  dp[u][depth] = a[u];
  for (int v : adj[u])
    if (v != p) DFS(v, u, depth + 1);
  for (int v : adj[u])
    if (v != p) {
      for (int i = 0; i <= n; i++) f[u][i] = 0;
      for (int i = depth; i <= n; i++)
        for (int j = depth + 1; j <= n; j++)
          if (i + j - 2 * depth > k)
            f[u][min(i, j)] = max(f[u][min(i, j)], dp[u][i] + dp[v][j]);
          else
            f[u][min(i, j)] = max(f[u][min(i, j)], dp[v][j]);
      for (int i = 0; i <= n; i++) dp[u][i] = max(dp[u][i], f[u][i]);
    }
}
void Solve() {
  DFS(1, -1, 0);
  ans = *max_element(dp[1], dp[1] + n);
  printf("%d", ans);
}
int main() {
  Read_Input();
  Solve();
  return 0;
}