"""

mod で場合分け
→使える数字の数が変わる

使える数字がX種類の時
数字の付け方を
2 <= X <= N の範囲で求めればいい？

1<=p<=Cの制約の影響を受けない範囲は省いてdp



"""

from sys import stdin
import sys
sys.setrecursionlimit(20000)

def dfs(v,p,num):

    if memo[v][num] != None:
        return memo[v][num]

    ret = 1
    for nex in lis[v]:
        tmp = 0
        if nex != p:
            if num+1 < c:
                tmp += dfs(nex,v,num+1)
            if num-1 >= 0:
                tmp += dfs(nex,v,num-1)
            ret *= tmp
            ret %= mod

    memo[v][num] = ret
    return ret
                

N,C = map(int,stdin.readline().split())
lis = [ [] for i in range(N) ]

for i in range(N-1):

    a,b = map(int,stdin.readline().split())
    a -= 1
    b -= 1
    lis[a].append(b)
    lis[b].append(a)

#mod 0
mod = 10**9+7
ans = 0

c = C//3
if c < 4*N:
    memo = [ [None] * c for i in range(N) ]
    for i in range(c):
        ans += dfs(0,0,i)
else:
    memo = [ [None] * (2*N) for i in range(N) ]
    for i in range(N):
        ans += dfs(0,0,i) * 2
    ans += pow(2,N-1,mod) * (c-2*N)

c = (C+2)//3
if c < 4*N:
    memo = [ [None] * c for i in range(N) ]
    for i in range(c):
        ans += dfs(0,0,i)
else:
    memo = [ [None] * (2*N) for i in range(N) ]
    for i in range(N):
        ans += dfs(0,0,i) * 2
    ans += pow(2,N-1,mod) * (c-2*N)

c = (C+1)//3
if c < 4*N:
    memo = [ [None] * c for i in range(N) ]
    for i in range(c):
        ans += dfs(0,0,i)
else:
    memo = [ [None] * (2*N) for i in range(N) ]
    for i in range(N):
        ans += dfs(0,0,i) * 2
    ans += pow(2,N-1,mod) * (c-2*N)

print (ans % mod)