human
What is a Python code to print the number of valid sets modulo 1000000007 given the following:
As you know, an undirected connected graph with n nodes and n - 1 edges is called a tree. You are given an integer d and a tree consisting of n nodes. Each node i has a value ai associated with it.

We call a set S of tree nodes valid if following conditions are satisfied:

  1. S is non-empty.
  2. S is connected. In other words, if nodes u and v are in S, then all nodes lying on the simple path between u and v should also be presented in S.
  3. .


Your task is to count the number of valid sets. Since the result can be very large, you must print its remainder modulo 1000000007 (109 + 7).

Input

The first line contains two space-separated integers d (0 ≤ d ≤ 2000) and n (1 ≤ n ≤ 2000).

The second line contains n space-separated positive integers a1, a2, ..., an(1 ≤ ai ≤ 2000).

Then the next n - 1 line each contain pair of integers u and v (1 ≤ u, v ≤ n) denoting that there is an edge between u and v. It is guaranteed that these edges form a tree.

Output

Print the number of valid sets modulo 1000000007.

Examples

Input

1 4
2 1 3 2
1 2
1 3
3 4

Output

8

Input

0 3
1 2 3
1 2
2 3

Output

3

Input

4 8
7 8 7 5 4 6 4 10
1 6
1 2
5 8
1 3
3 5
6 7
3 4

Output

41

Note

In the first sample, there are exactly 8 valid sets: {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {3, 4} and {1, 3, 4}. Set {1, 2, 3, 4} is not valid, because the third condition isn't satisfied. Set {1, 4} satisfies the third condition, but conflicts with the second condition.
from
assistant
#Here is Python code to print the number of valid sets modulo 1000000007.
collections import deque

class Graph:
  def __init__(self, n_vertices):
    self.n_vertices = n_vertices
    self.edges = [set() for i in xrange(n_vertices)]
    self.value = [-1] * n_vertices
    self.visit = [False]*n_vertices

  def color(self, i, c):
    if self.value[i] < 0:
      self.uncolored -= 1
      self.value[i] = c

      for neighbor in self.edges[i]:
        color(neighbor, 1-c)

  def set_value(self, i, val):
    self.value[i] = val

  def add_edge(self, i, j):
    self.edges[i].add(j)
    self.edges[j].add(i)

  def mydfs(self, u, lo, hi):
    self.visit[u] = True
    result = 1
    for v in self.edges[u]:
      if A[v] == lo and v<root: continue
      if not self.visit[v] and lo <= A[v] <= hi:
        result *= (self.mydfs(v,lo,hi)+1)
        result %= 10**9+7
    return result

  def dfs(self, start=1):
    stack = [start]
    discovered = [False]*self.n_vertices
    while stack:
      u = stack.pop()
      if not discovered[u]:
        yield u
        discovered[u] = True
        for neighbor in self.edges[u]:
          stack.append(neighbor)

  def bfs(self, start=1):
    stack = deque(start)
    discovered = [False]*self.n_vertices
    while stack:
      u = stack.popleft()
      if not discovered[u]:
        yield u
        discovered[u] = True
        for neighbor in self.edges[u]:
          stack.append(neighbor)

  def orderby(order_type='dfs'):
    if order_type == 'dfs':
      ordering = self.dfs()
    elif order_type == 'bfs':
      ordering = self.bfs()

    rank = [-1]*self.n_vertices
    for i, u in enumerate(ordering):
      rank[u] = i

    return rank

  def reverse_dfs(self, start=1):
    dfs_order = self.orderby(order_type='dfs')

    for u in reversed(list(self.dfs())):
      for child in self.edges[u]:
        if dfs_order[child] > dfs_order[u]:
          self.value[u] += self.value[child]

  def connected_components(self):
    remaining  = set(xrange(self.n_vertices))
    discovered = [False]*self.n_vertices

    while remaining:
      # get arbitrary start element
  ...