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 N = 2e5 + 5;
struct node {
  int u, v, w;
  bool operator<(const node &t) const { return w > t.w; }
} t;
int sz[N];
int pre[N];
long long ans[N];
priority_queue<node> q;
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
int find(int x) { return pre[x] == x ? x : pre[x] = find(pre[x]); }
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    if (i != n) q.push({read(), read(), read()});
    pre[i] = i;
    sz[i] = 1;
  }
  while (!q.empty()) {
    t = q.top(), q.pop();
    int u = t.u, fu = find(u);
    int v = t.v, fv = find(v);
    int w = t.w;
    ans[w] += 1ll * sz[fu] * sz[fv];
    pre[fu] = fv;
    sz[fv] += sz[fu];
  }
  for (int i = 1; i < N; i++) {
    ans[i] += ans[i - 1];
  }
  while (m--) {
    printf("%lld ", ans[read()]);
  }
}