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 N = 202;
int n, k;
int b[N];
vector<int> a[N];
int dp[N][N];
int ndp[N];
void dfs(int x, int p) {
  dp[x][0] = b[x];
  for (int i = 0; i < a[x].size(); ++i) {
    int h = a[x][i];
    if (h == p) continue;
    dfs(h, x);
    memset(ndp, 0, sizeof ndp);
    for (int h1 = 0; h1 <= n; ++h1) {
      for (int h2 = 0; h2 <= n; ++h2) {
        if (h1 + h2 + 1 <= k) continue;
        ndp[min(h1, h2 + 1)] = max(ndp[min(h1, h2 + 1)], dp[x][h1] + dp[h][h2]);
      }
    }
    memcpy(dp[x], ndp, sizeof ndp);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 1);
  int ans = 0;
  for (int i = 0; i <= n; ++i) ans = max(ans, dp[1][i]);
  printf("%d\n", ans);
  return 0;
}