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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
const long long mod = 1000000007;
const long long inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.000000000000000000001;
using namespace std;
long long n, k;
long long T[205], DP[205][205];
vector<long long> P[205];
void dfs(long long n, long long p) {
  DP[n][0] = T[n];
  for (auto v : P[n]) {
    if (v != p) dfs(v, n);
    DP[n][0] += DP[v][k - 1];
  }
  for (long long i = 1; i < k; i++) {
    long long maks = 0;
    long long suma = 0;
    for (auto v : P[n])
      if (v != p) {
        suma += DP[v][max(i - 1, k - i - 1)];
        maks = max(maks, DP[v][i - 1] - DP[v][max(i - 1, k - i - 1)]);
      }
    suma += maks;
    DP[n][i] = suma;
  }
  for (long long i = k - 2; i >= 0; i--) DP[n][i] = max(DP[n][i], DP[n][i + 1]);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  k++;
  for (long long i = 1; i <= n; i++) cin >> T[i];
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    P[a].push_back(b);
    P[b].push_back(a);
  }
  dfs(1, 0);
  cout << DP[1][0];
  return 0;
}