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 int N = 1e5 + 5;
const int MOD = 1e5 + 5;
int n, m, canh, tplt = n, cha[1000005];
long long kq[1000005], ans;
struct que {
  int u, v, w;
} b[1000005];
unordered_map<int, int> mapp;
struct query {
  int val, i;
} c[1000005];
bool operator<(query a, query b) { return a.val < b.val; }
bool operator<(que a, que b) { return a.w < b.w; }
int tim(int u) {
  if (cha[u] == 0) return u;
  return cha[u] = tim(cha[u]);
}
void join(int u, int v) {
  int x = tim(u);
  int y = tim(v);
  if (x == y) return;
  tplt--;
  if (x > y)
    cha[x] = y, ans += ((long long)(mapp[y] + 1) * (mapp[x] + 1)),
    mapp[y] += mapp[x] + 1, mapp[x] = -1;
  else
    cha[y] = x, ans += ((long long)(mapp[y] + 1) * (mapp[x] + 1)),
    mapp[x] += mapp[y] + 1, mapp[y] = -1;
}
void noi(int i) {
  while (b[canh + 1].w <= c[i].val && canh < n - 1) {
    canh++;
    int u = b[canh].u;
    int v = b[canh].v;
    join(u, v);
  }
}
void solve() {
  for (int i = 1; i <= m; i++) {
    int ts = c[i].val;
    noi(i);
    kq[c[i].i] = ans;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    b[i] = {u, v, w};
  }
  tplt = n;
  sort(b + 1, b + n);
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    c[i] = {x, i};
  }
  sort(c + 1, c + m + 1);
  solve();
  for (int i = 1; i <= m; i++) cout << kq[i] << " ";
  return 0;
}