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;
typedef struct edge {
  int u, v, w;
} edge;
edge E[200005];
long long ans[200005] = {0};
int p[200005];
long long sz[200005] = {0};
int comp(const void *a, const void *b) {
  return ((edge *)a)->w - ((edge *)b)->w;
}
int findSet(int u) {
  if (p[u] == u) return u;
  return p[u] = findSet(p[u]);
}
long long Union(int u, int v) {
  int pu = findSet(u);
  int pv = findSet(v);
  long long ans = 0;
  if (pu != pv) {
    if (sz[pu] <= sz[pv]) {
      p[pu] = pv;
      ans = (sz[pu] * (sz[pu] - 1)) / 2;
      ans += (sz[pv] * (sz[pv] - 1)) / 2;
      ans = -ans;
      sz[pv] = sz[pu] + sz[pv];
      sz[pu] = 0;
      ans += (sz[pv] * (sz[pv] - 1)) / 2;
    } else {
      p[pv] = pu;
      ans = (sz[pu] * (sz[pu] - 1)) / 2;
      ans += (sz[pv] * (sz[pv] - 1)) / 2;
      ans = -ans;
      sz[pu] = sz[pu] + sz[pv];
      sz[pv] = 0;
      ans += (sz[pu] * (sz[pu] - 1)) / 2;
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  int u, v, w;
  int mx = -1;
  for (int i = 0; i < n - 1; i++) {
    cin >> E[i].u >> E[i].v >> E[i].w;
    mx = max(mx, E[i].w);
  }
  qsort(E, n - 1, sizeof(edge), comp);
  for (int i = 0; i <= n; i++) {
    p[i] = i;
    sz[i] = 1;
  }
  int curwt = -1;
  for (int i = 0; i < n - 1; i++) {
    if (E[i].w != curwt && i > 0) {
      for (int j = E[i - 1].w + 1; j <= E[i].w; j++) ans[j] = ans[j - 1];
    }
    long long tmp = Union(E[i].u, E[i].v);
    ans[E[i].w] += tmp;
    curwt = E[i].w;
  }
  for (int i = 0; i < m; i++) {
    int x;
    cin >> x;
    if (x >= mx) {
      long long N = n;
      ans[x] = (long long)(N * (N - 1)) / 2;
    }
    cout << ans[x] << " ";
  }
  cout << endl;
  return 0;
}