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() {
  int n, m;
  cin >> n >> m;
  int s1 = 0, s2 = 0;
  int f = 0;
  vector<int> a, b;
  while (s1 < n || s2 < n) {
    int x, y;
    if (f == 0) {
      x = 9;
      y = 1;
      f = 1;
    } else if (f == 1) {
      x = 1;
      y = 8;
      f = 2;
    } else if (f == 2) {
      x = 8;
      y = 1;
      f = 1;
    }
    s1 += x;
    s2 += y;
    a.push_back(x);
    b.push_back(y);
  }
  for (int i = a.size() - 1; i >= 0; i--) cout << a[i];
  cout << endl;
  for (int i = b.size() - 1; i >= 0; i--) cout << b[i];
  return 0;
}
