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>
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const long long INF = 1LL << 60;
const long long MOD = 1000000007LL;
const long long MAX = 500000LL;
using namespace std;
long long cnt[200500];
class UnionFind {
 public:
  vector<long long> Parent;
  UnionFind(long long N) { Parent = vector<long long>(N, -1); }
  long long root(long long A) {
    if (Parent[A] < 0) {
      return A;
    }
    return Parent[A] = root(Parent[A]);
  }
  long long size(long long A) { return -Parent[root(A)]; }
  bool connect(long long A, long long B) {
    A = root(A);
    B = root(B);
    if (A == B) {
      return false;
    }
    if (size(A) < size(B)) {
      swap(A, B);
    }
    Parent[A] += Parent[B];
    Parent[B] = A;
    return true;
  }
};
int main() {
  long long n, m;
  scanf("%lld %lld", &n, &m);
  UnionFind uni(n);
  vector<tuple<long long, long long, long long>> vt(n - 1);
  for (long long i = 0; i < n - 1; i++) {
    long long x, y, z;
    scanf("%lld %lld %lld", &x, &y, &z);
    x--;
    y--;
    vt[i] = make_tuple(z, x, y);
  }
  sort(vt.begin(), vt.end());
  long long id = 0;
  for (long long i = 1; i <= 200000; i++) {
    cnt[i] += cnt[i - 1];
    while (id < vt.size()) {
      long long z, x, y;
      tie(z, x, y) = vt[id];
      if (i == z) {
        if (uni.root(x) != uni.root(y)) {
          cnt[i] -= uni.size(x) * (uni.size(x) - 1) / 2;
          cnt[i] -= uni.size(y) * (uni.size(y) - 1) / 2;
          uni.connect(x, y);
          cnt[i] += uni.size(x) * (uni.size(x) - 1) / 2;
        }
        id++;
      } else {
        break;
      }
    }
  }
  for (long long loop = 0; loop < m; loop++) {
    long long q;
    scanf("%lld", &q);
    printf("%lld", cnt[q]);
    if (loop == m - 1) {
      printf("\n");
    } else {
      printf(" ");
    }
  }
  return 0;
}