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;
template <class T>
void printArray(const T* a, int n, ostream& out = cerr);
template <class T>
void printArray(const vector<T>& arr, ostream& out = cerr);
const long long mod = 1e9 + 7;
const double PI = acos(-1);
int n, i, j, k, t;
int a[202];
vector<int> adj[202];
int dp[202][202];
int dfs(int v, int p, int rem) {
  int& ans = dp[v][rem];
  if (ans == -1) {
    ans = 0;
    if (rem == 0) {
      int tmp = a[v];
      for (int x : adj[v]) {
        if (x == p) continue;
        tmp += dfs(x, v, k);
      }
      ans = max(ans, tmp);
    }
    if (rem > (k + 1) / 2) {
      int tmp = 0;
      for (int x : adj[v]) {
        if (x == p) continue;
        tmp += dfs(x, v, max(rem - 1, 0));
      }
      ans = max(ans, tmp);
    }
    for (int i = max(1, rem); i <= (k + 1) / 2; ++i) {
      for (int x : adj[v]) {
        if (x == p) continue;
        int tmp = 0;
        for (int y : adj[v]) {
          if (y == p) continue;
          if (x == y) {
            tmp += dfs(y, v, i - 1);
          } else {
            tmp += dfs(y, v, k - i);
          }
        }
        ans = max(ans, tmp);
      }
    }
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int ans = 0;
  memset(dp, -1, sizeof dp);
  ans = max(ans, dfs(0, 0, 0));
  printf("%d\n", ans);
  return 0;
}
template <class T>
void printArray(const T* a, int n, ostream& out) {
  for (int i = 0; i < n; ++i) {
    out << a[i] << " ";
  }
  out << endl;
}
template <class T>
void printArray(const vector<T>& arr, ostream& out) {
  for (const T& x : arr) {
    out << x << " ";
  }
  out << endl;
}