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;
int n, m, par[200003];
long long child[200003], ans[200003];
struct query {
  int weight, id;
};
query q[200003];
struct edge {
  int weight, u, v;
};
edge g[200003];
bool cmpquery(const query& a, const query& b) { return (a.weight < b.weight); }
bool cmpedge(const edge& a, const edge& b) { return (a.weight < b.weight); }
int FindRoot(int u) {
  if (par[u] < 0) {
    return u;
  }
  return par[u] = FindRoot(par[u]);
}
void Uniion(int r, int s) {
  if (par[s] < par[r]) {
    swap(r, s);
  }
  par[r] += par[s];
  child[r] += child[s];
  par[s] = r;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    par[i] = -1, child[i] = 1;
  }
  for (int i = 1; i <= n - 1; ++i) {
    int weight, u, v;
    scanf("%d%d%d", &u, &v, &weight);
    g[i] = {weight, u, v};
  }
  for (int i = 1; i <= m; ++i) {
    int weight;
    scanf("%d", &weight);
    q[i] = {weight, i};
    ans[i] = 0;
  }
  sort(g + 1, g + 1 + n - 1, cmpedge);
  sort(q + 1, q + 1 + m, cmpquery);
  int idx = 1;
  q[0] = {0, 0};
  for (int i = 1; i <= m; ++i) {
    while (idx <= n - 1 && g[idx].weight <= q[i].weight) {
      int r = FindRoot(g[idx].u);
      int s = FindRoot(g[idx].v);
      if (r != s) {
        ans[q[i].id] += child[r] * child[s];
        Uniion(r, s);
      }
      ++idx;
    }
    ans[q[i].id] += ans[q[i - 1].id];
  }
  for (int i = 1; i <= m; ++i) {
    printf("%lld ", ans[i]);
  }
  return 0;
}