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 dp[maxn][maxn];
int tmp[maxn];
int n, k;
int a[maxn];
vector<int> vec[maxn];
void dfs(int u, int fa) {
  int v;
  for (int i = 0; i < vec[u].size(); i++) {
    v = vec[u][i];
    if (v == fa) continue;
    dfs(v, u);
    for (int x = 0; x <= 200; x++)
      for (int y = 0; y <= 200; y++) {
        if (x + y + 1 <= k) continue;
        tmp[min(x, y + 1)] = max(tmp[min(x, y + 1)], dp[u][x] + dp[v][y]);
      }
    for (int x = 0; x <= 200; x++) {
      dp[u][x] = tmp[x];
      tmp[x] = 0;
    }
  }
  int Max = 0;
  for (int i = k + 1; i <= 200; i++) Max = max(Max, dp[u][i]);
  dp[u][0] = a[u] + Max;
}
int main() {
  scanf("%d%d", &n, &k);
  int u, v;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  dfs(1, 0);
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= 200; j++) ans = max(ans, dp[i][j]);
  printf("%d", ans);
  return 0;
}