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 = 200 * 1000 + 10;
int n, ind, p[maxn], u, v, w, q, m, z[maxn];
long long cur, ans[maxn];
vector<pair<int, int>> qu;
vector<pair<int, pair<int, int>>> vec;
int gp(int x) {
  if (p[x] == x) return x;
  return p[x] = gp(p[x]);
}
void mrg(int a, int b) {
  a = gp(a);
  b = gp(b);
  if (z[b] < z[a]) swap(a, b);
  cur += (long long)z[a] * z[b];
  z[b] += z[a];
  p[a] = b;
}
int main() {
  ios::sync_with_stdio(0);
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    p[i] = i;
    z[i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d %d", &u, &v, &w);
    u--;
    v--;
    vec.push_back({w, {u, v}});
  }
  sort(vec.begin(), vec.end());
  for (int i = 0; i < m; i++) {
    scanf("%d", &q);
    qu.push_back({q, i});
  }
  sort(qu.begin(), qu.end());
  for (auto g : qu) {
    while (ind < n - 1 && vec[ind].first <= g.first) {
      mrg(vec[ind].second.first, vec[ind].second.second);
      ind++;
    }
    ans[g.second] = cur;
  }
  for (int i = 0; i < m; i++) printf("%lld ", ans[i]);
  return 0;
}