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>
char s1[333], s2[333];
int main() {
  s1[330] = '5';
  s2[330] = '5';
  for (int i = 329; i >= 0; i--)
    if (i % 2) {
      s1[i] = '5';
      s2[i] = '4';
    } else {
      s1[i] = '4';
      s2[i] = '5';
    }
  int n, m;
  scanf("%d%d", &n, &m);
  printf("%s\n%s\n", s1, s2);
  return 0;
}
