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 double eps = 1e-10;
const double Pi = 3.1415926535897932;
const int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1};
const int maxn = 2e5 + 5;
struct Edge {
  int a, b, w;
  bool operator<(const Edge &oth) const { return w < oth.w; }
} edges[maxn];
int n, m;
int uset[maxn];
int siz[maxn];
struct node {
  int q;
  int pos;
  bool operator<(const node &oth) const { return q < oth.q; }
} q[maxn];
long long ans[maxn];
int find(int x) {
  if (x != uset[x]) uset[x] = find(uset[x]);
  return uset[x];
}
inline long long calc(long long s) { return s * (s - 1) / 2; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n - 1; i++)
    scanf("%d%d%d", &edges[i].a, &edges[i].b, &edges[i].w);
  for (int i = 1; i <= m; i++) scanf("%d", &q[i].q), q[i].pos = i;
  ;
  sort(edges + 1, edges + 1 + n - 1);
  sort(q + 1, q + 1 + m);
  for (int i = 1; i <= n; i++) uset[i] = i, siz[i] = 1;
  int p = 1;
  long long now = 0;
  for (int i = 1; i <= m; i++) {
    while (p <= n - 1) {
      if (edges[p].w <= q[i].q) {
        Edge &e = edges[p];
        if (find(e.a != e.b)) {
          now -= calc(siz[find(e.a)]) + calc(siz[find(e.b)]);
          siz[find(e.a)] += siz[find(e.b)];
          uset[find(e.b)] = find(e.a);
          now += calc(siz[find(e.a)]);
        }
        p++;
      } else
        break;
    }
    ans[q[i].pos] = now;
  }
  for (int i = 1; i <= m; i++)
    printf("%lld", ans[i]), putchar(i == m ? '\n' : ' ');
  return 0;
}