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 N = 200100;
int n, m, q, par[N], sz[N], cnt;
long long ans[N];
vector<pair<int, pair<int, int> > > edges;
vector<pair<int, int> > v;
map<int, int> mp;
int get(int u) { return par[u] == u ? u : par[u] = get(par[u]); }
void mrg(int u, int v) {
  u = get(u);
  v = get(v);
  if (par[u] == par[v]) return;
  if (rand() & 1) swap(u, v);
  par[v] = u;
  mp[sz[u]]--;
  mp[sz[v]]--;
  if (mp[sz[u]] == 0) mp.erase(sz[u]);
  if (mp[sz[v]] == 0) mp.erase(sz[v]);
  sz[u] += sz[v];
  mp[sz[u]]++;
}
int main() {
  scanf("%d%d", &n, &m);
  mp[1] = n;
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    edges.push_back({w, {u, v}});
  }
  sort(edges.begin(), edges.end());
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    v.push_back({x, i});
  }
  sort(v.begin(), v.end());
  for (auto X : v) {
    int val = X.first;
    int idx = X.second;
    for (; cnt < n - 1; cnt++) {
      if (val >= edges[cnt].first)
        mrg(edges[cnt].second.first, edges[cnt].second.second);
      else
        break;
    }
    long long curr = 0;
    for (auto x : mp) curr += 1LL * x.first * (x.first - 1) / 2 * x.second;
    ans[idx] = curr;
  }
  for (int i = 1; i <= m; i++) printf("%lld ", ans[i]);
}