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 INF = 1e9 + 7;
const int N = 2e2 + 10;
vector<int> v[N];
int dp[N][N];
int sum[N];
int a[N];
int n, k;
void dfs(int x, int fa = 0) {
  for (auto& y : v[x]) {
    if (y == fa) continue;
    dfs(y, x);
  }
  memset((sum), (0), sizeof(sum));
  for (auto& y : v[x]) {
    if (y == fa) continue;
    for (int i = 0; i <= k; i++) {
      sum[i] += dp[y][i];
    }
    dp[x][0] += dp[y][k];
  }
  dp[x][0] += a[x];
  for (auto& y : v[x]) {
    if (y == fa) continue;
    for (int i = 1; i <= k; i++) {
      int o = max(i - 1, k - i);
      dp[x][i] = max(dp[x][i], dp[y][i - 1] + sum[o] - dp[y][o]);
    }
  }
  for (int i = k; i >= 0; i--) dp[x][i] = max(dp[x][i], dp[x][i + 1]);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1);
  int ans = 0;
  for (int i = 0; i <= k; i++) {
    ans = max(ans, dp[1][i]);
  }
  printf("%d\n", ans);
  return 0;
}