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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const int N = 407;
const long long INF = 1e18 + 7;
int a[N], dp[N][N], mx[N][N], pr[N][N];
int n, k;
vector<int> g[N];
void dfs(int v = 1, int p = 0, int h = 1) {
  for (int to : g[v]) {
    if (to == p) continue;
    dfs(to, v, h + 1);
    int y = h + k, m = 0;
    dp[v][h] += pr[to][h + k + 1];
    for (int x = h + 1; x <= n; x++) {
      mx[v][x] = max(mx[v][x], dp[to][x] - pr[to][max(x, y)]);
      dp[v][x] += pr[to][max(x, y)];
      y--;
    }
  }
  dp[v][h] += a[v];
  int y = h + k;
  for (int x = h + 1; x <= n; x++) {
    dp[v][x] += mx[v][x];
    y--;
  }
  for (int i = n; i >= 1; i--) {
    pr[v][i] = max(pr[v][i + 1], dp[v][i]);
  }
}
int main() {
  cin >> n >> k;
  memset(mx, -0x3f3f3f3f, sizeof(mx));
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs();
  cout << pr[1][1];
}