//a_N = 1 + 100 + 100^2 + … + 100^(N-1) = (100^N - 1) / 99
//10^9 + 7で割った余り…フェルマーの小定理より(100^N - 1) * 99^(10^9 + 5) % (10^9 + 7)と値が一致
//a_11で割った余り…N % 11 == 0のとき0より、a_N % a_11 = a_(N % 11)

#include <iostream>
#define int long long
using namespace std;

int powmod(int a, int n, int mod) {
	if (n == 0) return 1;
	if (n % 2 == 0) return powmod((a * a) % mod, n / 2, mod) % mod;
	return (a * powmod(a, n - 1, mod)) % mod;
}

signed main() {
	int n;
	cin >> n;
	
	int p = 1000000007;
	int a = (powmod(100, n, p) + p - 1) % p;
	int b = powmod(99, p - 2, p);
	cout << (a * b) % p << endl;
	
	int ans = 0, add = 1;
	for (int i = 0; i < n % 11; i++) {
		ans += add;
		add *= 100;
	}
	cout << ans << endl;
	
	return 0;
}