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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int INF = 1e9 + 1233;
const long long LINF = 1e18 + 1233;
template <typename T>
void print(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << ' ';
}
template <typename T>
void input(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
vector<int> p(2 * 100000 + 10);
vector<long long> sz(2 * 100000 + 10);
int find_set(int v) {
  if (v == p[v]) return v;
  return p[v] = find_set(p[v]);
}
void union_set(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (sz[a] < sz[b]) swap(a, b);
    p[b] = a;
    sz[a] += sz[b];
  }
}
struct edge {
  int v, u, w;
};
struct query {
  int idx;
  int q;
};
bool comp(edge a, edge b) { return a.w < b.w; }
bool comp2(query a, query b) { return a.q < b.q; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i < p.size(); i++) p[i] = i, sz[i] = 1;
  int n, m;
  cin >> n >> m;
  vector<edge> e(n - 1);
  for (int i = 0; i < n - 1; i++) cin >> e[i].u >> e[i].v >> e[i].w;
  vector<query> q(m);
  for (int i = 0; i < m; i++) cin >> q[i].q, q[i].idx = i;
  long long res = 0;
  sort((e).begin(), (e).end(), comp);
  sort((q).begin(), (q).end(), comp2);
  int pos = 0;
  vector<long long> ans(m);
  for (int i = 0; i < m; i++) {
    while (pos < n - 1 && e[pos].w <= q[i].q) {
      long long szu = sz[find_set(e[pos].u)];
      long long szv = sz[find_set(e[pos].v)];
      union_set(e[pos].u, e[pos].v);
      res = res - szu * (szu - 1) / 2 - szv * (szv - 1) / 2 +
            (szu + szv) * (szu + szv - 1) / 2;
      pos++;
    }
    ans[q[i].idx] = res;
  }
  for (auto to : ans) cout << to << ' ';
}