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;
bool check(long long n, long long pos) { return n & (1LL << pos); }
long long Set(long long n, long long pos) { return n = n | (1LL << pos); }
vector<long long> graph[203];
long long arr[203];
long long dp[203][203];
long long n, k;
void dfs(long long node, long long par = 0) {
  for (auto it : graph[node]) {
    if (it == par) continue;
    dfs(it, node);
  }
  for (long long d = 0; d <= n; d++) {
    if (d == 0) {
      dp[node][d] = arr[node];
      for (auto it : graph[node]) {
        if (it == par) continue;
        dp[node][d] += dp[it][k - 1];
      }
    } else {
      for (auto it : graph[node]) {
        if (it == par) continue;
        long long x = dp[it][d - 1];
        for (auto jt : graph[node]) {
          if (jt == par) continue;
          if (it == jt) continue;
          x += dp[jt][max(d - 1, k - d - 1)];
        }
        dp[node][d] = max(dp[node][d], x);
      }
    }
  }
  for (long long i = n - 1; i >= 0; i--) {
    dp[node][i] = max(dp[node][i], dp[node][i + 1]);
  }
}
int main() {
  long long i, j, l, m, o, r, q;
  long long testcase;
  long long input, flag, tag, ans;
  while (cin >> n >> k) {
    for (i = 1; i <= n; i++) graph[i].clear();
    k++;
    memset(dp, 0, sizeof dp);
    for (i = 1; i <= n; i++) scanf("%lld", &arr[i]);
    for (i = 1; i < n; i++) {
      long long u, v;
      scanf("%lld", &u);
      scanf("%lld", &v);
      graph[u].push_back(v);
      graph[v].push_back(u);
    }
    dfs(1);
    printf("%lld\n", dp[1][0]);
  }
}