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;
const int MAX = 1e5;
const ll oo = 1e15;
struct edge {
  int u, v, w;
  edge(int u, int v, int w) : u(u), v(v), w(w) {}
};
int n, m;
vector<pair<int, int>> q;
vector<edge> edges;
vector<ll> ans;
vector<ll> sz;
vector<int> p;
int find(int v) {
  if (p[v] == v) {
    return v;
  }
  return (p[v] = find(p[v]));
}
void unite(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) {
    return;
  }
  if (sz[u] > sz[v]) {
    swap(u, v);
  }
  p[u] = v;
  sz[v] += sz[u];
}
void solve() {
  cin >> n >> m;
  q.assign(m, {0, 0});
  edges.assign(n - 1, edge(0, 0, 0));
  p.assign(n + 1, 0);
  sz.assign(n + 1, 0);
  ans.assign(m, 0ll);
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    edges[i] = edge(u, v, w);
  }
  for (int i = 0; i < m; ++i) {
    int v;
    cin >> v;
    q[i] = make_pair(v, i);
  }
  if (n > 1) {
    for (int i = 1; i <= n; ++i) {
      p[i] = i;
      sz[i] = 1ll;
    }
    sort(edges.begin(), edges.end(),
         [](const edge &a, const edge &b) { return a.w < b.w; });
    sort(q.begin(), q.end());
    int b = 0;
    for (int i = 0; i < m; ++i) {
      int cq = q[i].first;
      ll cur = (i ? ans[q[i - 1].second] : 0ll);
      int j;
      for (j = b; j < n - 1 && edges[j].w <= cq; ++j) {
        int u = edges[j].u;
        int v = edges[j].v;
        u = find(u);
        v = find(v);
        if (u != v) {
          cur -= sz[u] * (sz[u] - 1) / 2;
          cur -= sz[v] * (sz[v] - 1) / 2;
          unite(u, v);
          u = find(u);
          cur += sz[u] * (sz[u] - 1) / 2;
        }
      }
      b = j;
      ans[q[i].second] = cur;
    }
    sort(q.begin(), q.end(),
         [](const pair<int, int> &a, const pair<int, int> &b) {
           return a.second < b.second;
         });
  }
  for (int i = 0; i < m; ++i) {
    cout << ans[q[i].second] << ' ';
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}