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 long long maxn = 205;
inline long long read() {
  char c = getchar();
  long long t = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    t = (t << 3) + (t << 1) + (c ^ 48);
    c = getchar();
  }
  return t * f;
}
long long n, k, a[maxn];
struct edge {
  long long v, p;
} e[maxn << 1];
long long h[maxn], cnt;
inline void add(long long a, long long b) {
  e[++cnt].p = h[a];
  e[cnt].v = b;
  h[a] = cnt;
  e[++cnt].p = h[b];
  e[cnt].v = a;
  h[b] = cnt;
}
long long dp[maxn][maxn];
void dfs(long long u, long long fa) {
  dp[u][0] = a[u];
  for (long long i = h[u]; i; i = e[i].p) {
    long long v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
  }
  for (long long dep = 0; dep < n; dep++) {
    if (dep == 0) {
      for (long long i = h[u]; i; i = e[i].p) {
        long long v = e[i].v;
        if (v == fa) continue;
        dp[u][dep] += dp[v][max(0ll, k - dep - 1)];
      }
    } else {
      for (long long i = h[u]; i; i = e[i].p) {
        long long v = e[i].v;
        if (v == fa) continue;
        long long cur = dp[v][dep - 1];
        for (long long i = h[u]; i; i = e[i].p) {
          long long v2 = e[i].v;
          if (v2 == v || v2 == fa) continue;
          cur += dp[v2][max(dep - 1, k - dep - 1)];
        }
        dp[u][dep] = max(dp[u][dep], cur);
      }
    }
  }
  for (long long dep = n - 1; dep > 0; dep--) {
    dp[u][dep - 1] = max(dp[u][dep], dp[u][dep - 1]);
  }
}
signed main() {
  n = read(), k = read();
  k++;
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i < n; i++) {
    long long a = read(), b = read();
    add(a, b);
  }
  dfs(1, 0);
  printf("%lld\n", dp[1][0]);
  return 0;
}