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;
long long int sumOfDigits(long long int n) {
  long long int s = 0;
  while (n) {
    s += (n % 10);
    n /= 10;
  }
  return s;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, s, i;
  string a, b;
  cin >> n >> m;
  a = '9';
  s = 9;
  while (s < n) {
    a += '9';
    s += 9;
  }
  b = a;
  for (i = 1; i < a.size(); i++) b += "0";
  b += '1';
  cout << a << ' ' << b;
  return 0;
}
