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;
int main() {
  long long i, j, k, l, m, n, s1 = 0, o = -1, p = -1, q = 0, mm = 0;
  cin >> n >> m;
  string s = "1", s2 = "9";
  k = 1;
  l = 9;
  while (l < n || k < n) {
    if (q % 2 == 0) {
      s2 = '0' + s2;
      s = '9' + s;
      k += 9;
    } else {
      s2 = '9' + s2;
      l += 9;
      s = '0' + s;
    }
    q++;
  }
  if (s[0] == '0') {
    for (i = 1; i < s.length(); i++) cout << s[i];
  } else
    cout << s;
  cout << " ";
  if (s2[0] == '0') {
    for (i = 1; i < s2.length(); i++) cout << s2[i];
  } else
    cout << s2;
}
