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(3)
using namespace std;
template <class T>
inline bool scan(T& ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' and (c < '0' or c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' and c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
template <class T>
inline void out(T x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
int _ = 0;
void testcase() { cout << "Case " << (++_) << ": "; }
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const double PI = acos(-1.0);
const long double eps = 1e-6;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
const int M = 710;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
struct E {
  int u, v, w;
} e[N];
int f[N];
int cnt[N];
int faf(int x) {
  if (x == f[x]) return x;
  return f[x] = faf(f[x]);
}
long long ans = 0;
void Union(int x, int y) {
  x = faf(x);
  y = faf(y);
  if (x == y) return;
  ans -= (long long)cnt[x] * (cnt[x] - 1) / 2;
  ans -= (long long)cnt[y] * (cnt[y] - 1) / 2;
  cnt[y] += cnt[x];
  f[x] = y;
  ans += (long long)cnt[y] * (cnt[y] - 1) / 2;
}
long long res[N];
void work() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    f[i] = i;
    cnt[i] = 1;
  }
  for (int i = 0; i < n - 1; ++i) {
    cin >> e[i].u >> e[i].v >> e[i].w;
  }
  sort(e, e + n - 1, [&](const E& lhs, const E& rhs) { return lhs.w < rhs.w; });
  for (int i = 0; i < n - 1; ++i) {
    Union(e[i].u, e[i].v);
    res[e[i].w] = ans;
  }
  for (int i = 1; i <= 2e5; ++i) {
    res[i] = max(res[i], res[i - 1]);
  }
  for (int i = 0; i < m; ++i) {
    int x;
    cin >> x;
    cout << res[x] << " ";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  work();
  return 0;
}