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 maxN = 205;
struct Node {
  int to, value, next;
} edge[maxN * 2 + 1];
int head[maxN + 1], tot;
int n, k, a[maxN + 1];
int f[maxN + 1][maxN + 1];
inline void add(int x, int y) {
  tot++;
  edge[tot].to = y;
  edge[tot].next = head[x];
  head[x] = tot;
}
inline void dfs(int u, int pa) {
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == pa) continue;
    dfs(v, u);
    for (int i = 1; i <= n; i++) {
      int val = f[u][i];
      for (int j = 0; j <= n; j++)
        if (i + j + 1 > k)
          f[u][min(i, j + 1)] = max(f[u][min(i, j + 1)], val + f[v][j]);
    }
  }
  f[u][0] = a[u];
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == pa) continue;
    f[u][0] += f[v][k];
  }
  for (int i = n; i >= 0; i--) f[u][i] = max(f[u][i], f[u][i + 1]);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  printf("%d", f[1][0]);
  return 0;
}