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;
template <typename T>
inline void read(T& x) {
  int w = 1;
  T data = 0;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
  x = data * w;
}
template <typename T>
void write(T x) {
  if (x < 0) putchar('-'), x = ~x + 1;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
inline void writeln(T x) {
  write(x), putchar('\n');
}
signed main() {
  int n, m;
  read(n), read(m);
  int four = n / 4, rem = n % 4 + 1, lst = 10 - rem;
  for (int i = 1; i <= four; ++i) putchar('4');
  putchar(rem + '0');
  putchar('\n');
  for (int i = 1; i <= four; ++i) putchar('5');
  putchar(lst + '0');
  putchar('\n');
  return 0;
}
