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. <image>.



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.
#import sys
#sys.stdin = open("in.txt",'r')
#sys.stdout = open("out.txt",'w')

d,n = map(int, raw_input().split(' '))
A = map(int, raw_input().split(" "))
G = [[] for i in range(n)]
for i in range(n-1):
    u, v = map(int, raw_input().split(' '))
    u -= 1
    v -= 1
    G[u].append(v)
    G[v].append(u)
M = 1000000007
flag = [[False for j in range(n)] for i in range(n)]
def cal(x, p, u):
    ret = 1
    for v in G[u]:
        if v != p and A[v] >= A[x] and A[v]-A[x] <= d and not flag[x][v]:
            flag[x][v] = flag[v][x] = True
            ret *= (1 + cal(x, u, v))
            ret %= M
    return ret
ans = 0
for u in range(n):
    x = cal(u, u, u)
    #print(x)
    ans += x
    ans %= M

print(ans)