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;
struct edge {
  int fr, to, we;
  bool operator<(const edge& rhs) const { return we < rhs.we; }
};
int n, m;
vector<edge> eds;
long long ans = 0;
int par[200005], sz[200005];
int fnd(int u) { return par[u] = (u == par[u] ? u : fnd(par[u])); }
bool unn(int u, int v) {
  u = fnd(u), v = fnd(v);
  par[u] = v;
  ans -= (1LL * sz[u] * (sz[u] - 1)) / 2;
  ans -= (1LL * sz[v] * (sz[v] - 1)) / 2;
  sz[v] += sz[u];
  ans += (1LL * sz[v] * (sz[v] - 1)) / 2;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int a, b, w, i = 1; i < n; i++) {
    scanf("%d%d%d", &a, &b, &w);
    eds.push_back({a, b, w});
  }
  sort(eds.begin(), eds.end());
  fill(sz + 1, sz + n + 1, 1);
  iota(par, par + n + 1, 0);
  vector<pair<int, long long>> res{{0, 0}};
  for (edge e : eds) {
    unn(e.fr, e.to);
    if (!res.empty() && e.we == res.back().first) res.pop_back();
    res.push_back({e.we, ans});
  }
  for (int q, i = 0; i < m; i++) {
    scanf("%d", &q);
    printf("%lld ",
           prev(upper_bound(res.begin(), res.end(), make_pair(q, LLONG_MAX)))
               ->second);
  }
  printf("\n");
}