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;
inline int rd() {
  int x = 0;
  char ch, t = 0;
  while (!isdigit(ch = getchar())) t |= ch == '-';
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return x = t ? -x : x;
}
int n, k, tot, ans;
int a[10005], hea[10005], tmp[10005];
int ver[10005 * 2], nex[10005 * 2];
int dp[10005][205];
void add(int x, int y) { ver[++tot] = y, nex[tot] = hea[x], hea[x] = tot; }
void dfs(int x, int fa) {
  for (int i = hea[x]; i; i = nex[i]) {
    if (ver[i] == fa) continue;
    dfs(ver[i], x);
    for (int j = 0; j <= k; j++) tmp[j] = -0x3f3f3f3f;
    for (int j = 1; j <= k; j++)
      for (int l = 0; l <= k; l++) {
        if (j + l + 1 < k) continue;
        tmp[min(j, l + 1)] = max(tmp[min(j, l + 1)], dp[x][j] + dp[ver[i]][l]);
      }
    for (int j = 1; j <= k; j++) dp[x][j] = max(dp[x][j], tmp[j]);
  }
  dp[x][0] = a[x] + dp[x][k];
  for (int i = k - 1; i >= 0; i--) dp[x][i] = max(dp[x][i], dp[x][i + 1]);
}
int main() {
  n = rd(), k = rd() + 1;
  for (int i = 1; i <= n; i++) a[i] = rd();
  int u, v;
  for (int i = 1; i < n; i++) {
    u = rd(), v = rd();
    add(u, v), add(v, u);
  }
  dfs(1, 0);
  printf("%d\n", dp[1][0]);
  return 0;
}