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 NMAX = 2e5 + 5;
const long long INF = 1e18;
struct al {
  long long id, val;
} q[NMAX];
struct edge {
  long long from, to, cost;
} edges[NMAX];
long long tata[NMAX], h[NMAX], rez[NMAX];
long long cnt;
long long findSet(long long node) {
  if (node != tata[node]) tata[node] = findSet(tata[node]);
  return tata[node];
}
void unionSet(long long p1, long long p2) {
  long long t1 = findSet(p1), t2 = findSet(p2);
  if (t1 == t2) return;
  cnt += h[t1] * h[t2];
  if (h[t1] > h[t2]) {
    h[t1] += h[t2];
    tata[t2] = t1;
  } else {
    h[t2] += h[t1];
    tata[t1] = t2;
  }
}
void addtotree(edge p1) { unionSet(p1.to, p1.from); }
bool cmp(const al &p1, const al &p2) { return p1.val < p2.val; }
bool cmp2(const edge &p1, const edge &p2) { return p1.cost < p2.cost; }
int main() {
  long long N, Q, i, j;
  cin >> N >> Q;
  for (i = 1; i < N; ++i) {
    long long x, y, c;
    cin >> x >> y >> c;
    edges[i].from = x;
    edges[i].to = y;
    edges[i].cost = c;
  }
  for (i = 1; i <= Q; ++i) {
    cin >> q[i].val, q[i].id = i;
  }
  for (i = 1; i <= N; ++i) tata[i] = i, h[i] = 1;
  sort(q + 1, q + Q + 1, cmp);
  sort(edges + 1, edges + N, cmp2);
  for (i = 1, j = 0; i <= Q; ++i) {
    while (j < N && edges[j + 1].cost <= q[i].val) {
      addtotree(edges[j + 1]);
      j++;
    }
    rez[q[i].id] = cnt;
  }
  for (i = 1; i <= Q; ++i) cout << rez[i] << " ";
  return 0;
}