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 mem[205][205];
int arr[205];
vector<int> adj[205];
int n, k;
int go(int node, int w, int pa = 0) {
  if (mem[node][w] != -1) return mem[node][w];
  mem[node][w] = 0;
  if (w == 0) {
    int cur = arr[node];
    for (int i : adj[node])
      if (i != pa) cur += go(i, k, node);
    mem[node][w] = max(mem[node][w], cur);
  }
  for (int x = max(w - 1, 0); x <= k; x++) {
    int nx = max(k - x - 1, 0);
    if (adj[node].size() > 2) nx = max(nx, k / 2);
    nx = max(nx, w - 1);
    int cur = 0, cur1 = 0;
    for (int i : adj[node])
      if (i != pa) cur1 += go(i, nx, node);
    for (int i : adj[node]) {
      if (i != pa) {
        int aa = cur1 - go(i, nx, node), bb = go(i, x, node);
        int z = cur1 - go(i, nx, node) + go(i, x, node);
        cur = max(cur, z);
      }
    }
    mem[node][w] = max(mem[node][w], cur);
  }
  return mem[node][w];
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  memset(mem, -1, sizeof mem);
  cout << go(1, 0) << endl;
}