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 n, m;
int A[MAXN], B[MAXN], C[MAXN];
int W[MAXN];
int d[MAXN], e[MAXN], f[MAXN], sz[MAXN];
long long ans[MAXN];
int Find(int x) { return (f[x] == x) ? x : (f[x] = Find(f[x])); }
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) f[i] = i, sz[i] = 1;
  for (int i = 1; i < n; ++i) {
    d[i] = i;
    scanf("%d%d%d", A + i, B + i, C + i);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", W + i);
    e[i] = i;
  }
  sort(d + 1, d + n, [&](int x, int y) { return C[x] < C[y]; });
  sort(e + 1, e + m + 1, [&](int x, int y) { return W[x] < W[y]; });
  long long sum = 0;
  for (int i = 1, j = 1; i <= m; ++i) {
    int x = e[i];
    while (j < n && C[d[j]] <= W[x]) {
      int y = d[j];
      sum += 1LL * sz[Find(A[y])] * sz[Find(B[y])];
      sz[Find(A[y])] += sz[Find(B[y])];
      f[Find(B[y])] = Find(A[y]);
      ++j;
    }
    ans[x] = sum;
  }
  for (int i = 1; i <= m; ++i) printf("%lld ", ans[i]);
  return 0;
}