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 = 210;
int n, k;
int a[maxn], d[maxn], dp[maxn][maxn], tmp[maxn];
int tot, head[maxn], nxt[maxn << 1], to[maxn << 1];
void add(int x, int y) {
  to[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
void dfs(int x, int f) {
  dp[x][0] = a[x];
  for (int i = head[x]; i; i = nxt[i]) {
    int y = to[i];
    if (y != f) dfs(y, x);
  }
  for (int i = 0; i < n; i++) {
    if (!i) {
      for (int l = head[x]; l; l = nxt[l]) {
        int y = to[l];
        if (y == f) continue;
        dp[x][0] += dp[y][k];
      }
    } else {
      for (int l = head[x]; l; l = nxt[l]) {
        int y = to[l];
        if (y == f) continue;
        int now = dp[y][i - 1];
        for (int j = head[x]; j; j = nxt[j]) {
          int yy = to[j];
          if (yy == f || yy == y) continue;
          now += dp[yy][max(i - 1, k - i)];
        }
        dp[x][i] = max(dp[x][i], now);
      }
    }
  }
  for (int i = n; i >= 1; i--) dp[x][i - 1] = max(dp[x][i - 1], dp[x][i]);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    add(u, v);
    add(v, u);
  }
  dfs(0, -1);
  printf("%d", dp[0][0]);
  return 0;
}