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 = 1 << 18;
pair<int, pair<int, int> > p[N];
struct node {
  int cnt, fa;
} f[N];
long long ans[N];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1, u, v, w; i < n; ++i) {
    scanf("%d%d%d", &u, &v, &w);
    p[i] = {w, {u, v}};
  }
  sort(p + 1, p + n);
  for (int i = 1; i <= n; ++i) f[i] = {1, i};
  function<int(int)> Find = [&](int x) {
    return x == f[x].fa ? x : f[x].fa = Find(f[x].fa);
  };
  function<long long(int, int)> mer = [&](int u, int v) {
    u = Find(u), v = Find(v);
    if (u == v) return 0LL;
    long long ret = 1LL * f[u].cnt * f[v].cnt;
    f[v].fa = u;
    f[u].cnt += f[v].cnt;
    return ret;
  };
  long long tp = 0;
  int np = 1;
  for (int now = 1, nw; now < n;) {
    nw = p[now].first;
    while (now < n && p[now].first == nw) {
      int u = p[now].second.first, v = p[now].second.second;
      tp += mer(u, v);
      now++;
    }
    while (np < nw) ans[np] = ans[np - 1], ++np;
    ans[np++] = tp;
  }
  while (np < N) ans[np] = ans[np - 1], ++np;
  for (int i = 0, x; i < m; ++i) {
    scanf("%d", &x);
    printf("%lld%c", ans[x], " \n"[i == m - 1]);
  }
}