def kaibun(y):
	x = len(str(y))
	flag = 0
	X = [9,9]
	i = 2
	while x > i:
		if i % 2 == 0:
			X.append(X[i - 1] * 10)
		else:
			X.append(X[i - 1])
		i = i + 1
	a = sum(X[:x - 1])
	for j in range(1, x // 2 + 1):
		if j == 1:
			a = a + (int(str(y)[j - 1]) - 1) * X[x - 2 * j + 1]//9
		else:
			a = a + (int(str(y)[j - 1])) * X[x - 2 * j + 1] // 9
	if y <= int(str(y)[::-1]):
		a = a + 1
	if x % 2 == 1:
		a = a + int(str(y)[x // 2])
	return a 
	# X[i] は i 桁の自然数のうち回文となるものの個数
n = int(input())

print(kaibun(n)%1000000000)
print(kaibun(n)%1000000007)