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 long long int N = 200000 + 5;
long long int par[N];
long long int C2(long long int x) {
  if (x < 2) {
    return 0;
  } else {
    return ((x * (x - 1)) / 2);
  }
}
long long int P(long long int x) {
  if (par[x] <= -1) {
    return x;
  } else {
    return par[x] = P(par[x]);
  }
}
void Union(long long int x, long long int y) {
  long long int px = P(x);
  long long int py = P(y);
  if (px == py) {
    return;
  } else {
    if (par[px] < par[py]) {
      par[px] = par[px] + par[py];
      par[py] = px;
    } else {
      par[py] = par[py] + par[px];
      par[px] = py;
    }
  }
}
class Edge {
 public:
  long long int start;
  long long int end;
  long long int weight;
};
bool cmp(Edge a, Edge b) { return a.weight < b.weight; }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  memset(par, -1, sizeof(par));
  long long int n, m;
  cin >> n >> m;
  if (n == 1) {
    for (long long int i = 0; i < m; i++) {
      cout << 0 << " ";
    }
    exit(0);
  }
  Edge e[n];
  e[n - 1].weight = 1000000000000;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> e[i].start >> e[i].end >> e[i].weight;
  }
  long long int A[m];
  for (long long int i = 0; i < m; i++) cin >> A[i];
  sort(e, e + n - 1, cmp);
  vector<pair<long long int, long long int> > ans(n - 1);
  ans[0].first = e[0].weight;
  ans[0].second = 1;
  Union(e[0].start, e[0].end);
  for (long long int i = 1; i < n - 1; i++) {
    ans[i].first = e[i].weight;
    long long int s = e[i].start;
    long long int ee = e[i].end;
    ans[i].second =
        ans[i - 1].second - C2(-1 * par[P(s)]) - C2(-1 * par[P(ee)]);
    Union(s, ee);
    ans[i].second += C2(-1 * par[P(s)]);
  }
  for (long long int i = 0; i < m; i++) {
    long long int x = A[i];
    long long int l = 0;
    long long int r = n - 2;
    long long int aaa = 0;
    while (l <= r) {
      long long int mid = (l + r) / 2;
      if (ans[mid].first <= x) {
        aaa = max(aaa, ans[mid].second);
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    cout << aaa << " ";
  }
}