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>
int n, m, Fa[200005], Siz[200005];
struct Edge {
  int x, y, z;
} e[200005];
inline bool operator<(const Edge& a, const Edge& b) { return a.z < b.z; }
struct Query {
  int q, ID;
} q[200005];
inline bool operator<(const Query& a, const Query& b) { return a.q < b.q; }
long long Ans[200005], Cur;
int Get(const int x) { return x == Fa[x] ? x : Fa[x] = Get(Fa[x]); }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; ++i) scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].z);
  for (int i = 1; i <= n; ++i) Siz[Fa[i] = i] = 1;
  for (int i = 1; i <= m; ++i) scanf("%d", &q[i].q), q[i].ID = i;
  std::sort(e + 1, e + n), std::sort(q + 1, q + m + 1);
  int Ae = 1;
  for (int i = 1; i <= m; ++i) {
    while (Ae < n && e[Ae].z <= q[i].q) {
      int x = e[Ae].x, y = e[Ae].y;
      ++Ae, x = Get(x), y = Get(y);
      if (x == y) continue;
      Cur -= Siz[x] * (Siz[x] - 1ll) >> 1;
      Cur -= Siz[y] * (Siz[y] - 1ll) >> 1;
      Siz[x] += Siz[y], Fa[y] = x;
      Cur += Siz[x] * (Siz[x] - 1ll) >> 1;
    }
    Ans[q[i].ID] = Cur;
  }
  for (int i = 1; i <= m; ++i) printf("%I64d%c", Ans[i], i == m ? '\n' : ' ');
  return 0;
}