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);
  int i, j, k;
  int n, m;
  cin >> n >> m;
  int a[10000], b[10000], c[10000] = {0};
  int q = n / 9, r = n % 9;
  int ca = 0, cb = 0;
  for (i = 1; i <= q; i++) {
    a[ca++] = 9;
    b[cb++] = 9;
  }
  if (r != 0) a[ca++] = r;
  b[cb++] = r;
  b[cb++] = 10 - r;
  for (i = 0; i < ca; i++) {
    cout << a[i];
  }
  cout << " ";
  ca = ca - 1;
  c[ca] = 10 - a[ca];
  for (i = ca - 1; i >= 0; i--) {
    c[i] = 10 - (a[i] + 1);
  }
  for (i = 1; i <= 150; i++) {
    cout << "9";
  }
  for (i = 0; i <= ca; i++) {
    cout << c[i];
  }
  cout << "\n";
  return 0;
}