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 = 205;
int dp[N][N], sum[N][N], n, k, a[N], ans;
vector<int> V[N];
void dfs(int u, int p) {
  for (int j = 0; j <= n; j++) sum[u][j] = dp[u][j] = 0;
  for (int i = 0; i < V[u].size(); i++) {
    if (V[u][i] == p) continue;
    dfs(V[u][i], u);
    for (int j = 0; j <= n; j++) {
      sum[u][j] += dp[V[u][i]][j];
      if (j) dp[u][j] = max(dp[u][j], dp[V[u][i]][j - 1]);
    }
  }
  for (int j = n; j >= 1; j--) {
    dp[u][j] = max(dp[u][j], dp[u][j + 1]);
  }
  for (int Mn = 0; Mn <= (k - 1) / 2; Mn++) {
    for (int j = 0; j < V[u].size(); j++) {
      if (V[u][j] == p) continue;
      dp[u][Mn + 1] =
          max(dp[u][Mn + 1],
              sum[u][k - Mn - 1] - dp[V[u][j]][k - Mn - 1] + dp[V[u][j]][Mn]);
    }
  }
  for (int Mn = (k - 1) / 2 + 1; Mn <= n; Mn++) {
    for (int j = 0; j < V[u].size(); j++) {
      if (V[u][j] == p) continue;
      dp[u][Mn + 1] =
          max(dp[u][Mn + 1], sum[u][Mn] - dp[V[u][j]][Mn] + dp[V[u][j]][Mn]);
    }
  }
  for (int j = n; j >= 1; j--) {
    dp[u][j] = max(dp[u][j], dp[u][j + 1]);
    sum[u][j] = sum[u][j - 1];
  }
  sum[u][0] = a[u];
  dp[u][0] = max(dp[u][1], a[u] + dp[u][k + 1]);
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 2; i <= n; i++) {
    int u, v;
    cin >> u >> v;
    V[u].push_back(v);
    V[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    dfs(i, 0);
    ans = max(ans, dp[i][0]);
  }
  cout << ans;
}