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 = 2e5 + 5;
int n, m;
int pre[N], sz[N];
int find_set(int x) {
  if (x == pre[x]) return x;
  return pre[x] = find_set(pre[x]);
}
struct node {
  int u, v, w;
} p[N];
pair<int, int> q[N];
bool cmp(node a, node b) { return a.w < b.w; }
long long ans[N];
int main() {
  cin >> n >> m;
  for (int i = (1); i < (n); i++) {
    cin >> p[i].u >> p[i].v >> p[i].w;
  }
  for (int i = (1); i < (m + 1); i++) {
    cin >> q[i].first;
    q[i].second = i;
  }
  for (int i = (1); i < (n + 1); i++) pre[i] = i, sz[i] = 1;
  sort(q + 1, q + m + 1);
  sort(p + 1, p + n, cmp);
  long long cnt = 0, cur = 1;
  for (int i = 1; i <= m; ++i) {
    while (cur < n && p[cur].w <= q[i].first) {
      int fx = find_set(p[cur].u);
      int fy = find_set(p[cur].v);
      cnt += (long long)sz[fx] * sz[fy];
      pre[fx] = fy;
      sz[fy] += sz[fx];
      cur++;
    }
    ans[q[i].second] = cnt;
  }
  for (int i = 1; i <= m; ++i) cout << ans[i] << " ";
  return 0;
}