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;
int n, k, a[400], t, linkk[400], f[401][401];
struct hzy {
  int val, next;
} e[1001];
void ling(int x, int y) {
  e[++t].val = y;
  e[t].next = linkk[x];
  linkk[x] = t;
}
void dfs(int x, int fa) {
  f[x][0] = a[x];
  for (int i = linkk[x]; i; i = e[i].next) {
    int y = e[i].val;
    if (y != fa) {
      dfs(y, x);
    }
    f[x][0] += f[y][k];
  }
  for (int t = 1; t < n; t++) {
    for (int i = linkk[x]; i; i = e[i].next) {
      int y = e[i].val;
      if (y != fa) {
        int s = f[y][t - 1];
        for (int other = linkk[x]; other; other = e[other].next) {
          int yy = e[other].val;
          if (yy != fa && yy != y) {
            s += f[yy][max(t - 1, k - t)];
          }
        }
        f[x][t] = max(f[x][t], s);
      }
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    f[x][i - 1] = max(f[x][i - 1], f[x][i]);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    ling(x, y);
    ling(y, x);
  }
  dfs(1, 0);
  printf("%d", f[1][0]);
}