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 = 2e5 + 5;
long long p[N], sz[N], res[N];
vector<int> a[N], b[N];
int f(int r) { return p[r] != r ? p[r] = f(p[r]) : r; }
long long g(int u, int v) {
  int x = f(u), y = f(v);
  long long res = sz[x] * sz[y];
  p[x] = y;
  sz[y] += sz[x];
  return res;
}
int main() {
  int n, m, x, y, z;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) p[i] = i, sz[i] = 1;
  for (int i = 1; i < n; i++) {
    cin >> x >> y >> z;
    a[z].push_back(x);
    b[z].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    res[i] = res[i - 1];
    for (int j = 0; j < a[i].size(); j++) res[i] += g(a[i][j], b[i][j]);
  }
  while (m--) {
    cin >> x;
    cout << res[x] << " ";
  }
}