Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, x = 0, y = 0, r = 0, l = 0, m;
  string s = "", s1 = "", s3 = "";
  cin >> n >> m;
  while (x < n) {
    x += 9;
    s += (9 + '0');
    s += '0';
  }
  s += '9';
  s1 += '9';
  int dd = s.size();
  int sum = 0;
  for (int i = 0; i < dd; i++) {
    if (i == dd - 1)
      s1 += '1';
    else if (s[i] == '0')
      s1 += '9';
    else
      s1 += '0';
  }
  cout << s << "\n";
  cout << s1 << "\n";
  return 0;
}