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;
using pii = pair<int, int>;
using ll = long long;
using llu = long long unsigned;
using ld = long double;
const ld EPS = 1e-9;
inline int cmp(ld first, ld second = 0, ld tol = EPS) {
  return (first <= second + tol) ? (first + tol < second) ? -1 : 0 : 1;
}
const int MOD = 1;
inline int mod(ll first, int m = MOD) { return (int)(((first % m) + m) % m); }
const int N = 2e5 + 10;
int n, m;
int sz[N], link[N];
vector<pii> edges[N];
ll arr[N];
int find(int first) {
  if (link[first] == first) return first;
  return link[first] = find(link[first]);
}
bool same(int first, int second) { return find(first) == find(second); }
void unite(int u, int v) {
  u = find(u), v = find(v);
  if (sz[u] > sz[v]) swap(u, v);
  sz[v] += sz[u];
  link[u] = v;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) sz[i] = 1, link[i] = i;
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    edges[w].push_back({u, v});
  }
  for (int w = 1; w <= 2e5; w++) {
    arr[w] = arr[w - 1];
    for (auto e : edges[w]) {
      if (!same(e.first, e.second)) {
        ll szNow = sz[find(e.first)];
        arr[w] -= (szNow * (szNow - 1)) / 2;
        szNow = sz[find(e.second)];
        arr[w] -= (szNow * (szNow - 1)) / 2;
        unite(e.first, e.second);
        szNow = sz[find(e.second)];
        arr[w] += (szNow * (szNow - 1)) / 2;
      }
    }
  }
  while (m--) {
    int q;
    scanf("%d", &q);
    printf("%lld ", arr[q]);
  }
  putchar('\n');
  return 0;
}