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 maxn = 2e2 + 10;
vector<int> G[maxn];
int n, K, a[maxn], dep[maxn], maxdep[maxn];
int dp[maxn][maxn], tmp[maxn];
void dfs(int u, int fa) {
  maxdep[u] = dep[u];
  dp[u][dep[u]] = a[u];
  for (int v : G[u])
    if (v != fa) {
      dep[v] = dep[u] + 1;
      dfs(v, u);
      maxdep[u] = max(maxdep[u], maxdep[v]);
    }
  for (int v : G[u])
    if (v != fa) {
      for (int i = dep[u]; i <= maxdep[u]; ++i) tmp[i] = dp[u][i];
      for (int i = dep[u]; i <= maxdep[u]; ++i)
        for (int j = dep[v]; j <= maxdep[v]; ++j)
          if (i + j - dep[u] * 2 > K)
            dp[u][min(i, j)] = max(dp[u][min(i, j)], tmp[i] + dp[v][j]);
      for (int i = dep[v]; i <= maxdep[v]; ++i)
        dp[u][i] = max(dp[u][i], dp[v][i]);
    }
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1, u, v; i < n; ++i) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dep[1] = 1;
  dfs(1, 0);
  int ans = 0;
  for (int i = dep[1]; i <= maxdep[1]; ++i) ans = max(ans, dp[1][i]);
  printf("%d\n", ans);
  return 0;
}