#! python3
# (prev * 10 + nextdigit) % 225
# only need to keep track of prev % x, where 10 * x % 225 = 0 (so prev * 10 % 225 and (prev + x) * 10 % 225 are the same)
# also need x % 10 == 0, so we can get the last digit
# x = 90 is smallest value that works

X = 90
mod = 1000000007

# create matrix
mat = [[0 for i in range(X)] for j in range(X)]
for prevmod in range(X):
	prevdig = prevmod % 10
	for nextdig in range(10):
		if nextdig == prevdig:
			continue
		mat[prevmod][(prevmod*10+nextdig)%X] += 1

# some methods for exponentiating matrices
def mult(a,b):
	ret = [[0 for i in range(X)] for j in range(X)]
	for i in range(X):
		for j in range(X):
			for k in range(X):
				ret[i][j] = (ret[i][j] + a[i][k] * b[k][j]) % mod
	return ret

def exp(mat,e):
	if e == 1:
		return mat
	if e % 2 == 1:
		return mult(mat, exp(mat, e - 1))
	tmp = mult(mat,mat)
	return exp(tmp, e//2)


# actual solution below
a,b = map(int, input().split())

if a == 1:
	print(1 if b < 10 else 0)
elif a == 2:
	print(1 if b < 100 and b % 11 != 0 and b >= 10 else 0)
else:
	mat = exp(mat, a - 2)
	ans = 0
	for startdig in range(1,10):
		for startmod in range(X):
			ways = mat[startdig][startmod]
			for enddig in range(10):
				endmod = (startmod * 10 + enddig) % 225
				if enddig != (startmod % 10) and endmod == b:
					ans += ways
	print(ans % mod)


# brute = 0
# for i in range(1000000):
# 	if len(str(i)) == a and i % 225 == b:
# 		s = str(i)
# 		for j in range(1,len(s)):
# 			if s[j] == s[j-1]:
# 				break
# 		else:
# 			brute += 1
# print(brute)