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;
int fastMax(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ y; }
int fastMin(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ x; }
const long long int MAXN = 206;
vector<long long int> graph[MAXN];
long long int a[MAXN];
long long int n, k;
long long int dp[MAXN][MAXN];
long long int depth[MAXN];
long long int tmp[MAXN];
void dfs(long long int u, long long int p) {
  depth[u] = 1;
  dp[u][0] = a[u];
  for (long long int v : graph[u]) {
    if (v == p) continue;
    dfs(v, u);
    long long int nw = max(depth[u], depth[v] + 1);
    for (long long int i = (0); i <= (nw - 1); ++i) tmp[i] = -1e9;
    for (long long int i = (0); i <= (depth[u] - 1); ++i) {
      for (long long int j = max(0LL, k - i); j < depth[v]; j++) {
        tmp[min(i, j + 1)] = max(tmp[min(i, j + 1)], dp[u][i] + dp[v][j]);
      }
    }
    for (long long int i = (0); i <= (depth[v] - 1); ++i)
      tmp[i + 1] = max(tmp[i + 1], dp[v][i]);
    for (long long int i = (0); i <= (depth[u] - 1); ++i)
      dp[u][i] = max(dp[u][i], tmp[i]);
    for (long long int i = (depth[u]); i <= (nw - 1); ++i) dp[u][i] = tmp[i];
    depth[u] = nw;
    for (long long int i = (depth[u] - 1); i >= (1); --i)
      dp[u][i - 1] = max(dp[u][i - 1], dp[u][i]);
  }
}
void solve() {
  cin >> n >> k;
  for (long long int i = (1); i <= (n); ++i) cin >> a[i];
  for (long long int i = (1); i <= (n - 1); ++i) {
    long long int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs(1, 0);
  cout << dp[1][0] << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  t = 1;
  for (long long int i = (1); i <= (t); ++i) {
    solve();
  }
}