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;
template <typename T, typename U>
inline void amin(T &a, U b) {
  a = (a > b ? b : a);
}
template <typename T, typename U>
inline void amax(T &a, U b) {
  a = (a > b ? a : b);
}
const int N = 2e5 + 5;
int f[N], sz[N];
int64_t ans = 0;
void init(int n) {
  for (int i = 0; i < n; ++i) f[i] = i, sz[i] = 1;
}
int root(int x) {
  while (x != f[x]) {
    f[x] = f[f[x]];
    x = f[x];
  }
  return x;
}
int64_t calc(int64_t x) {
  int64_t res = 0;
  amax(res, x * (x - 1) / 2);
  return res;
}
bool merge(int x, int y) {
  int a = root(x), b = root(y);
  if (a == b) return true;
  if (sz[a] > sz[b]) swap(a, b);
  ans -= calc(sz[a]);
  ans -= calc(sz[b]);
  f[a] = b;
  sz[b] += sz[a];
  ans += calc(sz[b]);
  return false;
}
bool isConnected(int x, int y) {
  if (root(x) == root(y))
    return true;
  else
    return false;
}
void solve() {
  int n, q;
  cin >> n >> q;
  init(n);
  vector<array<int, 3>> edges;
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    edges.push_back({w, u, v});
  }
  sort(edges.begin(), edges.end());
  do {
  } while (0);
  vector<pair<int, int>> query(q);
  for (int i = 0; i < q; ++i) {
    cin >> query[i].first;
    query[i].second = i;
  }
  sort(query.begin(), query.end());
  int L = 0;
  vector<int64_t> pairs(q, 0);
  for (auto &[W, i] : query) {
    do {
    } while (0);
    while (L < n - 1 && edges[L][0] <= W) {
      merge(edges[L][1], edges[L][2]);
      ++L;
    }
    do {
    } while (0);
    pairs[i] = ans;
  }
  for (auto i : pairs) cout << i << " ";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tests = 1;
  while (tests--) {
    solve();
  }
}