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;
struct newdata {
  int v, label;
};
struct doubly {
  int u, v, w;
};
int n, m, cnt;
int fa[200001];
long long sum[200001];
long long ans[200001];
newdata query[200001];
doubly path[200001];
bool cmp(newdata i, newdata j) { return i.v < j.v; }
bool cmp1(doubly i, doubly j) { return i.w < j.w; }
int find(int x) {
  if (fa[x] == 0)
    return x;
  else {
    fa[x] = find(fa[x]);
    return fa[x];
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    path[i].u = u;
    path[i].v = v;
    path[i].w = w;
  }
  for (int i = 1; i <= n; i++) sum[i] = 1;
  for (int i = 1; i <= m; i++) {
    query[i].label = i;
    scanf("%d", &query[i].v);
  }
  sort(query + 1, query + m + 1, cmp);
  sort(path + 1, path + n, cmp1);
  int now = 1;
  long long tmp = 0;
  for (int i = 1; i <= m; i++) {
    while (path[now].w <= query[i].v && now < n) {
      int x = find(path[now].u);
      int y = find(path[now].v);
      if (x != y) {
        tmp -= (sum[x] - 1) * sum[x] / 2;
        tmp -= (sum[y] - 1) * sum[y] / 2;
        fa[y] = x;
        sum[x] += sum[y];
        tmp += (sum[x] - 1) * sum[x] / 2;
      }
      now++;
    }
    ans[query[i].label] = tmp;
  }
  for (int i = 1; i <= m; i++) printf("%I64d ", ans[i]);
  return 0;
}