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>
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
int dxx[8] = {0, 0, 1, -1, -1, -1, 1, 1};
int dyy[8] = {1, -1, 0, 0, -1, 1, 1, -1};
using namespace std;
int par[200005];
long long int sz[200005];
long long int res = 0;
long long int calc(long long int x) { return x * 1ll * (x - 1) / 2; }
void intialize(int n) {
  for (long long int i = 0; i < n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
}
int find(int x) {
  if (par[x] != x) par[x] = find(par[x]);
  return par[x];
}
void merge(int x, int y) {
  int xr = find(x), yr = find(y);
  if (xr == yr) return;
  if (sz[xr] >= sz[yr]) {
    res -= calc(sz[xr]);
    res -= calc(sz[yr]);
    par[yr] = xr;
    sz[xr] += sz[yr];
    res += calc(sz[xr]);
  } else {
    res -= calc(sz[xr]);
    res -= calc(sz[yr]);
    par[xr] = yr;
    sz[yr] += sz[xr];
    res += calc(sz[yr]);
  }
}
int main() {
  intialize(200005);
  int n, m;
  cin >> n >> m;
  vector<pair<long long int, pair<int, int> > > v;
  for (long long int i = 0; i < n - 1; i++) {
    int x, y;
    long long int wt;
    cin >> x >> y >> wt;
    v.push_back({wt, {x, y}});
  }
  vector<pair<long long int, int> > q;
  vector<long long int> ans(m);
  for (long long int i = 0; i < m; i++) {
    long long int x;
    cin >> x;
    q.push_back({x, i});
  }
  sort(v.begin(), v.end());
  sort(q.begin(), q.end());
  int indx = 0;
  for (long long int i = 0; i < m; i++) {
    while (indx < n - 1 && v[indx].first <= q[i].first) {
      merge(v[indx].second.first, v[indx].second.second);
      indx++;
    }
    ans[q[i].second] = res;
  }
  for (long long int i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}