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 double EPS = 1e-9;
const double PI = acos(-1.0);
template <typename T>
inline T sq(T a) {
  return a * a;
}
template <typename T1, typename T2>
inline pair<T1, T2> mp(T1 a, T2 b) {
  return make_pair(a, b);
}
template <typename T1, typename T2>
inline T1 safeMod(T1 a, T2 m) {
  return (a % m + m) % m;
}
template <typename T1, typename T2>
inline bool isEq(T1 a, T2 b) {
  return abs(a - b) < EPS;
}
template <typename T1, typename T2, typename T3>
inline bool isEq(T1 a, T2 b, T3 eps) {
  return abs(a - b) < eps;
}
template <typename T>
inline bool isKthBitOn(T n, int k) {
  assert(n <= numeric_limits<T>::max());
  assert(k <= numeric_limits<T>::digits);
  T ONE = 1;
  return bool((n & (ONE << k)));
}
template <typename T>
inline void setKthBit(T& n, int k) {
  assert(n <= numeric_limits<T>::max());
  assert(k <= numeric_limits<T>::digits);
  T ONE = 1;
  n = (n | (ONE << k));
}
const int oo = 0x3f3f3f3f;
const int MAX = 200010;
const int MOD = 1000000007;
const int precision = 10;
vector<pair<int, pair<int, int>>> edges;
set<int> reps;
int comp[MAX], compsz[MAX];
int findrep(int n) {
  if (n == comp[n])
    return n;
  else
    return comp[n] = findrep(comp[n]);
}
long long sum = 0;
void makeUnion(int x, int y) {
  int a = findrep(x);
  int b = findrep(y);
  if (a != b) {
    comp[a] = b;
    sum -= (long long)compsz[a] * (compsz[a] - 1) / 2;
    sum -= (long long)compsz[b] * (compsz[b] - 1) / 2;
    reps.erase(a);
    compsz[b] += compsz[a];
    sum += (long long)compsz[b] * (compsz[b] - 1) / 2;
    if (compsz[b] > 1) reps.insert(b);
  }
}
long long ans[MAX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q, u, v, w;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    comp[i] = i;
    compsz[i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v >> w;
    edges.push_back({w, {u, v}});
  }
  sort((edges).begin(), (edges).end());
  for (int i = 0; i < n - 1; i++) {
    int j = i;
    set<int> vv;
    while (j < n - 1 && edges[i].first == edges[j].first) {
      tie(u, v) = edges[j].second;
      makeUnion(u, v);
      vv.insert(u);
      vv.insert(v);
      j++;
    }
    ans[edges[i].first] = sum;
    i = j - 1;
  }
  for (int i = 1; i < MAX; i++)
    if (ans[i] == 0) ans[i] = ans[i - 1];
  while (q--) {
    cin >> w;
    cout << ans[w] << " ";
  }
  return 0;
}