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;
int f[2000000];
long long sum[2000000];
long long ans[2000000];
long long temp = 0;
struct tr {
  int u;
  int v;
  int w;
};
tr t[2000000];
struct tr2 {
  int l;
  int r;
};
tr2 t2[2000000];
bool cmp(tr x, tr y) { return x.w < y.w; }
bool cmp1(tr2 x, tr2 y) { return x.l < y.l; }
int find(int x) {
  if (f[x] != x) f[x] = find(f[x]);
  return f[x];
}
void nion(int a, int b) {
  int x, y;
  x = find(a);
  y = find(b);
  if (x != y) {
    temp = temp - (sum[x] * (sum[x] - 1) / 2);
    temp = temp - (sum[y] * (sum[y] - 1) / 2);
    f[x] = y;
    sum[y] = sum[x] + sum[y];
    temp = temp + (sum[y] * (sum[y] - 1) / 2);
  }
}
int main() {
  int n, m;
  cin >> n;
  cin >> m;
  int i = 1;
  while (i < n) {
    scanf("%d%d%d", &t[i].u, &t[i].v, &t[i].w);
    i++;
  }
  i = 0;
  while (i < m) {
    scanf("%d", &t2[i].l);
    t2[i].r = i;
    i++;
  }
  i = 0;
  while (i <= n) {
    f[i] = i;
    sum[i] = 1;
    i++;
  }
  sort(t + 1, t + n, cmp);
  sort(t2, t2 + m, cmp1);
  int len = 1;
  for (i = 0; i < m; i++) {
    while (t2[i].l >= t[len].w && len < n) {
      nion(t[len].u, t[len].v);
      len++;
    }
    ans[t2[i].r] = temp;
  }
  i = 0;
  while (i < m) {
    printf("%lld ", ans[i]);
    i++;
  }
  printf("\n");
  return 0;
}