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;
const int N = 2 * 1e5 + 10;
long long res;
long long a[N];
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m;
  cin >> n >> m;
  string s;
  string s1;
  s = s + '8';
  s1 = s1 + '1';
  for (int i = 0; i < 150; ++i) {
    s = s + '9';
  }
  for (int i = 0; i < 150; ++i) {
    s = s + '0';
  }
  s = s + '1';
  for (int i = 0; i < 150; ++i) {
    s1 = s1 + '0';
  }
  for (int i = 0; i < 151; ++i) {
    s1 = s1 + '9';
  }
  cout << s << "\n";
  cout << s1;
  return 0;
}