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 = 2e5 + 7;
const int MAXV = 507;
const int MAXE = 507;
const int INF = 0x3f3f3f3f;
int n, m;
int par[MAXN];
int find(int x) {
  if (x == par[x]) return x;
  return par[x] = find(par[x]);
}
bool isSame(int x, int y) {
  int px = find(x), py = find(y);
  return px == py;
}
void unite(int x, int y) {
  int px = find(x), py = find(y);
  if (px == py) return;
  par[px] = py;
}
struct Edge {
  int u, v, cost;
  Edge() {}
  Edge(int u, int v, int cost) : u(u), v(v), cost(cost) {}
  bool operator<(Edge e) const { return cost < e.cost; }
} edge[MAXN];
int queryIdx[MAXN];
long long ans[MAXN];
long long cnt[MAXN];
bool vis[MAXN];
vector<int> qArray;
long long C2(long long x) { return x * (x - 1) / 2; }
void solve() {
  memset(vis, 0, sizeof(vis));
  sort(edge, edge + n - 1);
  sort(qArray.begin(), qArray.end());
  int now = 0, j = 0, k = 0;
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    cnt[i] = 1;
  }
  vector<long long> cv;
  int i;
  for (int j = 0; j < qArray.size(); j++) {
    for (i = k; i < n - 1; i++) {
      if (edge[i].cost > qArray[j]) {
        k = i;
        break;
      }
      int u = edge[i].u, v = edge[i].v;
      int pu = find(u), pv = find(v);
      res = res - C2(cnt[pu]) - C2(cnt[pv]);
      long long tmp = cnt[pu] + cnt[pv];
      cnt[pu] = cnt[pv] = tmp;
      res += C2(tmp);
      unite(u, v);
    }
    k = i;
    ans[qArray[j]] = res;
  }
  for (int i = 0; i < m; i++) {
    printf("%I64d ", ans[queryIdx[i]]);
  }
  putchar('\n');
}
int main() {
  while (cin >> n >> m) {
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < n - 1; i++) {
      int u, v, c;
      scanf("%d%d%d", &u, &v, &c);
      edge[i] = Edge(u, v, c);
    }
    for (int i = 0; i < m; i++) {
      int q;
      scanf("%d", &q);
      if (!vis[q]) {
        vis[q] = true;
        qArray.push_back(q);
      }
      queryIdx[i] = q;
    }
    solve();
  }
  return 0;
}