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;
struct edge {
  int v, n;
} e[400005];
int head[400005], k, dep[400005], a[400005], cnt, dp[405][405];
inline void add(int u, int v) {
  e[++cnt] = {v, head[u]};
  head[u] = cnt;
}
inline void dfs(int u, int f) {
  for (int i = head[u]; i; i = e[i].n) {
    if (e[i].v == f) continue;
    dep[e[i].v] = dep[u] + 1;
    dfs(e[i].v, u);
  }
  int dt[405] = {};
  for (int j = head[u]; j; j = e[j].n) {
    if (e[j].v == f) continue;
    for (int i = dep[u] + 1; i <= 400; i++) {
      if ((i - dep[u]) * 2 > k)
        dp[u][i] += dp[e[j].v][i];
      else {
        dp[u][i] += dp[e[j].v][(k + 1) - i + dep[u] + dep[u]],
            dt[i] = max(dt[i], dp[e[j].v][i] -
                                   dp[e[j].v][(k + 1 - i + dep[u]) + dep[u]]);
      }
    }
  }
  for (int i = 1; i <= 400; i++) dp[u][i] += dt[i];
  for (int i = dep[u] + k + 1; i <= 400; i++)
    dp[u][dep[u]] = max(dp[u][dep[u]], dp[u][i] + a[u]);
  for (int i = 400; i >= 0; i--) dp[u][i] = max(dp[u][i], dp[u][i + 1]);
}
int main(int argc, char** argv) {
  int n;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    add(u, v), add(v, u);
  }
  dfs(1, 0);
  cout << dp[1][0] << endl;
  return 0;
}