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>
const int mod = 998244353;
const int maxn = 1e6 + 5;
const int inf = 1e7;
const long long onf = 1e18;
using namespace std;
int n, m;
int q[maxn], root[maxn], sum[maxn];
long long ans[maxn];
struct node {
  int u, v, val;
  bool friend operator<(node a, node b) { return a.val < b.val; }
} a[maxn];
int Find(int x) {
  if (x == root[x]) return x;
  return root[x] = Find(root[x]);
}
void init() {
  for (int i = 1; i <= n; i++) {
    sum[i] = 1, root[i] = i;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  init();
  for (int i = 1; i < n; i++) scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].val);
  sort(a + 1, a + n);
  int End = 0;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &q[i]);
    End = max(End, q[i]);
  }
  for (int i = 1; i < n; i++) {
    int x1 = Find(a[i].u);
    int y1 = Find(a[i].v);
    long long temp = 0;
    if (x1 != y1) {
      temp += (long long)sum[x1] * sum[y1];
      sum[y1] += sum[x1], sum[x1] = 0;
      root[x1] = y1;
    }
    ans[a[i].val] += temp;
  }
  for (int i = 1; i <= End; i++) ans[i] += ans[i - 1];
  for (int i = 1; i <= m; i++) printf("%lld ", ans[q[i]]);
  return 0;
}