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 long long INF = 1e18;
const long long mod = 1e9 + 7;
const long long maxn = 1000 * 2001;
long long par[maxn], siz[maxn];
long long get_par(long long v) {
  while (v != par[v]) v = par[v];
  return v;
}
void dsu(long long v, long long u) {
  u = get_par(u);
  v = get_par(v);
  if (siz[u] > siz[v]) {
    par[v] = u;
    siz[u] = siz[v] + siz[u];
  } else {
    par[u] = v;
    siz[v] = siz[u] + siz[v];
  }
}
long long n, m, q;
vector<pair<long long, pair<long long, long long> > > edge;
long long ans[maxn], ww[maxn];
vector<long long> w;
int main() {
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    par[i] = i;
    siz[i] = 1;
  }
  for (long long i = 1; i < n; i++) {
    long long u, v, z;
    cin >> u >> v >> z;
    edge.push_back({z, {u, v}});
  }
  sort(edge.begin(), edge.end());
  long long anss = 0;
  for (long long i = 0; i < n - 1; i++) {
    long long u = (edge[i].second).first;
    long long v = (edge[i].second).second;
    long long z = edge[i].first;
    long long x = get_par(u);
    long long y = get_par(v);
    anss += ((siz[x] + siz[y]) * (siz[x] + siz[y] - 1)) / 2 -
            (siz[y] * (siz[y] - 1)) / 2 - (siz[x] * (siz[x] - 1)) / 2;
    dsu(u, v);
    ans[z] = max(anss, ans[z]);
  }
  long long x = 0;
  for (long long i = 1; i <= 2e5; i++)
    if (ans[i] == 0)
      ans[i] = x;
    else
      x = ans[i];
  for (long long i = 1; i <= q; i++) {
    long long qq;
    cin >> qq;
    ww[i] = qq;
  }
  for (long long i = 1; i <= q; i++) cout << ans[ww[i]] << " ";
}