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 long long MX = 210;
vector<long long> adj[MX];
long long used[MX][6 * MX];
long long dp[MX][6 * MX];
long long a[MX];
long long n;
long long h[MX];
long long dfs(long long u, long long p = -1) {
  for (long long v : adj[u]) {
    if (v != p) {
      h[u] = max(dfs(v, u), h[u]);
    }
  }
  h[u]++;
  return h[u];
}
const long long INF = 1e18;
const long long ADD = 500;
long long k;
long long f(long long u, long long p, long long d) {
  if (used[u][d + ADD]) {
    return dp[u][d + ADD];
  }
  used[u][d + ADD] = true;
  long long ans = 0;
  if (d <= 0) {
    ans = a[u];
    for (long long v : adj[u]) {
      if (v != p) {
        ans += f(v, u, k - 1);
      }
    }
  }
  for (long long x = 1; x <= h[u] - 1; x++) {
    long long sum = 0;
    long long maxi = -INF;
    for (long long v : adj[u]) {
      if (v != p) {
        if (h[v] >= x) {
          long long aux = max({k - x - 1, d - 1, x - 1});
          sum += f(v, u, aux);
          maxi = max(maxi, f(v, u, max(d - 1, x - 1)) - f(v, u, aux));
        }
      }
    }
    ans = max(ans, sum + maxi);
  }
  return dp[u][d + ADD] = ans;
}
void addEdge(long long u, long long v) {
  u--;
  v--;
  adj[u].push_back(v);
  adj[v].push_back(u);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  k++;
  for (long long i = 0; i < (long long)n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < (long long)n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    addEdge(u, v);
  }
  dfs(0);
  cout << f(0, -1, 0) << endl;
  return 0;
}