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 n, q;
struct Edge {
  int u, v, w;
  friend bool operator<(const Edge &x, const Edge &y) { return x.w < y.w; }
} e[200005];
long long ans[200005];
int fa[200005], sz[200005];
int fnd(int x) { return x == fa[x] ? x : fa[x] = fnd(fa[x]); }
long long unite(int x, int y) {
  x = fnd(x), y = fnd(y);
  if (x == y) return 0;
  long long now = 1LL * sz[x] * (sz[x] - 1) / 2 + 1LL * sz[y] * (sz[y] - 1) / 2;
  fa[y] = x, sz[x] += sz[y];
  return 1LL * sz[x] * (sz[x] - 1) / 2 - now;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; ++i) scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
  sort(e + 1, e + n);
  for (int i = 1; i <= n; ++i) fa[i] = i, sz[i] = 1;
  for (int i = 1; i < n; ++i) {
    int u = e[i].u, v = e[i].v, w = e[i].w;
    ans[w] += unite(u, v);
  }
  for (int i = 1; i <= n; ++i) ans[i] += ans[i - 1];
  while (q--) {
    int x;
    scanf("%d", &x);
    printf("%lld%c", ans[x], " \n"[!q]);
  }
}