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>
struct node {
  int a, b, w;
  bool operator<(const node &p) const { return w < p.w; }
} e[200002];
int w[200002], f[200002], n, m, top = 0;
long long ans[200002], s[200002];
inline int gf(int p) { return f[p] == p ? p : (f[p] = gf(f[p])); }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &e[i].a, &e[i].b, &e[i].w);
  }
  std::sort(e + 1, e + n);
  for (int i = 1; i <= n; i++) f[i] = i, s[i] = 1;
  for (int i = 1; i < n; i++) {
    if (e[i].w != e[i - 1].w) {
      w[++top] = e[i].w;
      ans[top] = ans[top - 1];
    }
    int x = gf(e[i].a), y = gf(e[i].b);
    ans[top] = ans[top] - (s[x] - 1) * s[x] - (s[y] - 1) * s[y] +
               (s[x] + s[y] - 1) * (s[x] + s[y]);
    f[x] = y;
    s[y] += s[x];
  }
  for (int i = 1; i <= top; i++) ans[i] /= 2;
  while (m--) {
    scanf("%d", &n);
    printf("%lld ", ans[std::upper_bound(w + 1, w + top + 1, n) - w - 1]);
  }
}