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>
int inp() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum;
}
int head[210], nxt[410], end[410];
long long f[210][210];
int cou = 0;
void link(int a, int b) {
  nxt[++cou] = head[a];
  head[a] = cou;
  end[cou] = b;
}
int n, k, a[210];
void dfs(int cur, int last) {
  f[cur][0] = a[cur];
  for (int x = head[cur]; x != -1; x = nxt[x])
    if (end[x] != last) {
      dfs(end[x], cur);
      f[cur][0] += f[end[x]][k];
    }
  for (int i = 1; i <= n; i++) {
    long long sum = 0;
    for (int x = head[cur]; x != -1; x = nxt[x])
      if (end[x] != last) sum += f[end[x]][std::max(i - 1, k - i)];
    for (int x = head[cur]; x != -1; x = nxt[x])
      if (end[x] != last)
        f[cur][i] =
            std::max(f[cur][i], sum - f[end[x]][std::max(i - 1, k - i)] +
                                    f[end[x]][i - 1]);
  }
  for (int i = n - 1; i >= 0; i--)
    f[cur][i] = std::max(f[cur][i], f[cur][i + 1]);
}
int main() {
  memset(head, -1, sizeof(head));
  n = inp();
  k = inp();
  for (int i = 1; i <= n; i++) a[i] = inp();
  for (int i = 1; i < n; i++) {
    int u = inp();
    int v = inp();
    link(u, v);
    link(v, u);
  }
  dfs(1, 0);
  printf("%lld\n", f[1][0]);
}