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 + 5;
int pre[maxn];
long long sum[maxn], ans[maxn];
int cnt = 0, n, m;
struct node1 {
  int u, v, w;
} e[maxn];
struct node2 {
  int id, val;
} ques[maxn];
int Find(int x) {
  int a = x;
  while (x != pre[x]) x = pre[x];
  while (a != pre[a]) {
    int temp = a;
    a = pre[a];
    pre[temp] = x;
  }
  return x;
}
bool cmp(node1 &n1, node1 &n2) { return n1.w < n2.w; }
bool cmp2(node2 &n1, node2 &n2) { return n1.val < n2.val; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i <= n; i++) pre[i] = i, sum[i] = 1;
  for (cnt = 0; cnt < n - 1; cnt++) {
    scanf("%d%d%d", &e[cnt].u, &e[cnt].v, &e[cnt].w);
  }
  sort(e, e + n - 1, cmp);
  for (int i = 0; i < m; i++) {
    scanf("%d", &ques[i].val);
    ques[i].id = i;
  }
  sort(ques, ques + m, cmp2);
  long long summ = 0;
  int t = 0;
  for (int i = 0; i < m; i++) {
    while (t < n - 1 && e[t].w <= ques[i].val) {
      int fu = Find(e[t].u);
      int fv = Find(e[t].v);
      if (fu == fv) {
        continue;
      } else {
        summ -= sum[fu] * (sum[fu] - 1) / 2;
        summ -= sum[fv] * (sum[fv] - 1) / 2;
        if (fv < fu) {
          pre[fu] = fv;
          sum[fv] += sum[fu];
          summ += (sum[fv] * (sum[fv] - 1) / 2);
        } else {
          pre[fv] = fu;
          sum[fu] += sum[fv];
          summ += (sum[fu] * (sum[fu] - 1) / 2);
        }
      }
      t++;
    }
    ans[ques[i].id] = summ;
  }
  for (int i = 0; i < m; i++) {
    printf("%I64d%c", ans[i], (i == m - 1) ? '\n' : ' ');
  }
  return 0;
}