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 = 300000;
int parent[N], size[N];
vector<pair<int, pair<int, int>>> v;
pair<int, int> q[N];
long long ans[N];
int Find(int a) {
  if (parent[a] == a) return a;
  return parent[a] = Find(parent[a]);
}
void Union(int a, int b) { parent[Find(b)] = Find(a); }
long long doit(int a, int b) {
  int x = Find(a);
  int y = Find(b);
  long long p = size[x];
  long long qq = size[y];
  long long anss = p * qq;
  size[x] += size[y];
  Union(x, y);
  return anss;
}
int main(int argc, char const *argv[]) {
  for (int i = 0; i < N; i++) parent[i] = i, size[i] = 1;
  int n, m, a, b, c;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d%d", &a, &b, &c);
    v.push_back({c, {a, b}});
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < m; i++) {
    scanf("%d", &a);
    q[i] = {a, i};
  }
  sort(q, q + m);
  int at = 0;
  long long sum = 0;
  for (int i = 0; i < m; i++) {
    if (at < v.size()) {
      while (v[at].first <= q[i].first) {
        if (at >= v.size()) break;
        long long kk = doit(v[at].second.first, v[at].second.second);
        sum += kk;
        at++;
        if (at >= v.size()) break;
      }
    }
    ans[q[i].second] = sum;
  }
  for (int i = 0; i < m; i++) {
    printf("%lld ", ans[i]);
  }
  return 0;
}