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 mx = 200100;
struct node {
  int x, y, z;
} a[mx];
struct nod {
  int id, x;
} b[mx];
int fa[mx];
long long num[mx];
long long ans[mx];
int n, m;
bool cmp(node x, node y) { return x.z < y.z; }
bool cmp1(nod x, nod y) { return x.x < y.x; }
int find(int x) {
  if (x != fa[x]) {
    return fa[x] = find(fa[x]);
  }
  return fa[x];
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) fa[i] = i, num[i] = 1;
  for (int i = 1; i < n; i++) {
    cin >> a[i].x >> a[i].y >> a[i].z;
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i].x;
    b[i].id = i;
  }
  sort(a + 1, a + n, cmp);
  sort(b + 1, b + m + 1, cmp1);
  int l = 1;
  long long sum = 0;
  for (int i = 1; i < n; i++) {
    while (l <= m && a[i].z > b[l].x) {
      ans[b[l].id] = sum;
      l++;
    }
    int x = find(a[i].x);
    int y = find(a[i].y);
    sum += num[y] * num[x];
    fa[x] = y;
    num[y] += num[x];
  }
  while (l <= m) {
    if (a[n - 1].z <= b[l].x) {
      ans[b[l].id] = sum;
    }
    l++;
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << " \n"[i == m];
  return 0;
}