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 u, v, c;
} e[200005];
struct Query {
  int val, id;
} q[200005];
int cnt[200005], p[200005];
long long ans, res[200005];
bool cmpE(Edge x, Edge y) { return x.c < y.c; }
bool cmpQ(Query x, Query y) { return x.val < y.val; }
int find(int x) {
  if (p[x] == x) return x;
  return (p[x] = find(p[x]));
}
void merge(int x, int y) {
  x = find(x);
  y = find(y);
  p[x] = y;
  ans -= 1LL * cnt[x] * (cnt[x] - 1) / 2;
  ans -= 1LL * cnt[y] * (cnt[y] - 1) / 2;
  cnt[y] = cnt[x] + cnt[y];
  ans += 1LL * cnt[y] * (cnt[y] - 1) / 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    cin >> e[i].u >> e[i].v >> e[i].c;
  }
  for (int i = 1; i <= n; i++) {
    p[i] = i;
    cnt[i] = 1;
  }
  e[n].c = 1e9;
  for (int i = 1; i <= m; i++) {
    cin >> q[i].val;
    q[i].id = i;
  }
  sort(q + 1, q + 1 + m, cmpQ);
  sort(e + 1, e + n, cmpE);
  int id = 1;
  ans = 0;
  for (int i = 1; i <= m; i++) {
    while (e[id].c <= q[i].val) {
      merge(e[id].u, e[id].v);
      id++;
    }
    res[q[i].id] = ans;
  }
  for (int i = 1; i <= m; i++) {
    cout << res[i] << ' ';
  }
  return 0;
}