Let s(x) be sum of digits in decimal representation of positive integer x. Given two integers n and m, find some positive integers a and b such that 

  * s(a) ≥ n, 
  * s(b) ≥ n, 
  * s(a + b) ≤ m. 

Input

The only line of input contain two integers n and m (1 ≤ n, m ≤ 1129).

Output

Print two lines, one for decimal representation of a and one for decimal representation of b. Both numbers must not contain leading zeros and must have length no more than 2230.

Examples

Input

6 5


Output

6 
7


Input

8 16


Output

35 
53

Note

In the first sample, we have n = 6 and m = 5. One valid solution is a = 6, b = 7. Indeed, we have s(a) = 6 ≥ n and s(b) = 7 ≥ n, and also s(a + b) = s(13) = 4 ≤ m.
#include <bits/stdc++.h>
using namespace std;
const double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230;
const long long MOD = 1e9 + 7;
const long long INF = 1e14;
long long mpow(long long a, long long b, long long p = MOD) {
  a = a % p;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p;
    b = b >> 1;
  }
  return res % p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  for (int i = 0; i < (int)(2000); ++i) cout << "5";
  cout << "\n";
  for (int i = 0; i < (int)(1999); ++i) cout << "4";
  cout << "5";
  return 0;
}
