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 = 210;
const int INF = 0x3f3f3f3f;
const int iinf = 1 << 30;
const long long linf = 2e18;
const int mod = 1000000007;
const double eps = 1e-7;
void douout(double x) { printf("%lf\n", x + 0.0000000001); }
template <class T>
void print(T a) {
  cout << a << endl;
  exit(0);
}
template <class T>
void chmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void Tmin(T &a, T &b, T &c) {
  return min(a, min(b, c));
}
template <class T>
void Tmax(T &a, T &b, T &c) {
  return max(a, max(b, c));
}
void add(int &a, int b) { a = a + b < mod ? a + b : a + b - mod; }
void sub(int &a, int b) { a = (a - b + mod) % mod; }
void mul(int &a, int b) { a = (long long)a * b % mod; }
int addv(int a, int b) { return (a += b) >= mod ? a -= mod : a; }
int subv(int a, int b) { return (a -= b) < 0 ? a += mod : a; }
int mulv(int a, int b) { return (long long)a * b % mod; }
int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int pw(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) s = (long long)s * a % mod;
  return s;
}
int n, k;
vector<int> e[N];
int a[N];
int dp[N][N];
void dfs(int x, int fa) {
  dp[x][0] = a[x];
  for (int i = (0); i <= (((int)e[x].size()) - 1); i++) {
    int to = e[x][i];
    if (to == fa) continue;
    dfs(to, x);
    dp[x][0] += dp[to][k];
  }
  for (int i = (1); i <= (n - 1); i++) {
    for (int j = (0); j <= (((int)e[x].size()) - 1); j++) {
      int to = e[x][j];
      if (to == fa) continue;
      int cnt = dp[to][i - 1];
      for (int o = (0); o <= (((int)e[x].size()) - 1); o++) {
        int oth = e[x][o];
        if (oth == to || oth == fa) continue;
        cnt += dp[oth][max(i - 1, k - i)];
      }
      chmax(dp[x][i], cnt);
    }
  }
  for (int i = (n - 1); i >= (0); i--) chmax(dp[x][i - 1], dp[x][i]);
}
signed main() {
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
  for (int i = (1); i <= (n - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, 0);
  printf("%d\n", dp[1][0]);
  return 0;
}