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;
using ll = long long;
using pii = pair<int, int>;
template <typename T>
using V = vector<T>;
struct edge {
  int u, v, w;
  bool operator<(const edge& other) const { return w < other.w; }
};
struct union_find {
  vector<int> parent;
  vector<int> size;
  int components = 0;
  ll curans = 0;
  union_find(int n = 0) {
    if (n > 0) init(n);
  }
  void init(int n) {
    parent.resize(n + 1);
    size.assign(n + 1, 1);
    components = n;
    for (int i = 0; i <= n; ++i) parent[i] = i;
  }
  int find(int v) { return v == parent[v] ? v : parent[v] = find(parent[v]); }
  bool unite(int u, int v) {
    u = find(u);
    v = find(v);
    if (u == v) return false;
    if (size[u] < size[v]) swap(u, v);
    curans += (ll)size[u] * size[v];
    parent[v] = u;
    size[u] += size[v];
    --components;
    return true;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<edge> edges(n - 1);
  vector<pii> q(m);
  vector<ll> ans(m, 0);
  for (int i = 0; i < n - 1; ++i) {
    cin >> edges[i].u >> edges[i].v >> edges[i].w;
  }
  sort(edges.begin(), edges.end());
  for (int i = 0; i < m; ++i) {
    cin >> q[i].first;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  union_find UF(n);
  int curp = 0;
  for (auto Q : q) {
    while (curp < n - 1 && edges[curp].w <= Q.first) {
      UF.unite(edges[curp].u, edges[curp].v);
      ++curp;
    }
    ans[Q.second] = UF.curans;
  }
  for (int i = 0; i < m; ++i) {
    cout << ans[i] << " \n"[i == m - 1];
  }
}