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 maxn = 2e5 + 10;
struct nod {
  int u, v, w;
} a[maxn];
bool cmp(nod a, nod b) { return a.w < b.w; }
int n, m, fa[maxn];
long long num[maxn], ans[maxn];
int find(int k) {
  if (k == fa[k]) return k;
  return fa[k] = find(fa[k]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i < n; i++) cin >> a[i].u >> a[i].v >> a[i].w;
  for (int i = 1; i <= n; i++) {
    fa[i] = i;
    num[i] = 1;
  }
  sort(a + 1, a + n, cmp);
  for (int i = 1; i < n; ++i) {
    int fu = find(a[i].u), fv = find(a[i].v);
    ans[a[i].w] += num[fu] * num[fv];
    num[fu] += num[fv];
    fa[fv] = fu;
  }
  for (int i = 1; i <= maxn; i++) ans[i] += ans[i - 1];
  int t;
  while (m--) {
    cin >> t;
    cout << ans[t] << " ";
  }
  return 0;
}