"""

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:
                if memo[nex][num+1] == None:
                    tmp += dfs(nex,v,num+1)
                else:
                    tmp += memo[nex][num+1]
            if num-1 >= 0:
                if memo[nex][num-1] == None:
                    tmp += dfs(nex,v,num-1)
                else:
                    tmp += memo[nex][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


clis = [C//3,C//3+1]
anss = []
for c in clis:
    ans = 0
    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)
    anss.append(ans)

cl2 = [ C//3 , (C+2)//3 , (C+1)//3 ]
aaa = 0
for i in cl2:
    for j in range(2):
        if clis[j] == i:
            aaa += anss[j]
print (aaa % mod)