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 mxn = 2e5 + 100;
int parent[mxn], sz[mxn];
long long ans[mxn], ret = 0;
array<int, 3> edges[mxn];
array<int, 2> queries[mxn];
int find(int p) {
  if (p == parent[p]) return p;
  return parent[p] = find(parent[p]);
}
long long nC2(int n) { return (1LL * n * (n - 1)) / 2; }
void merge(int u, int v) {
  int pu = find(u), pv = find(v);
  if (pu == pv) return;
  if (sz[pu] < sz[pv]) swap(pu, pv);
  ret += nC2(sz[pu] + sz[pv]) - nC2(sz[pu]) - nC2(sz[pv]);
  sz[pu] += sz[pv], parent[pv] = pu;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) parent[i] = i, sz[i] = 1;
  for (int i = 0; i < n - 1; i++)
    scanf("%d%d%d", &edges[i][1], &edges[i][2], &edges[i][0]);
  for (int i = 0; i < m; i++) scanf("%d", &queries[i][0]), queries[i][1] = i;
  sort(edges, edges + n - 1);
  sort(queries, queries + m);
  for (int j = 0, i = 0; j < m; ++j) {
    while (i < n - 1 && edges[i][0] <= queries[j][0])
      merge(edges[i][1], edges[i][2]), ++i;
    ans[queries[j][1]] = ret;
  }
  for (int i = 0; i < m; i++) printf("%lld ", ans[i]);
}