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 n, m;
vector<int> a, b;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  int i, j, k;
  int sum = 0;
  int suma = 5;
  int sumb = 5;
  a.push_back(5);
  b.push_back(5);
  for (i = 0; suma < n || sumb < n; ++i) {
    if (i % 2 == 0) {
      a.push_back(5);
      b.push_back(4);
      suma += 5;
      sumb += 4;
    } else {
      a.push_back(4);
      b.push_back(5);
      suma += 4;
      sumb += 5;
    }
  }
  for (i = a.size() - 1; i >= 0; --i) {
    cout << a[i];
  }
  cout << endl;
  for (i = b.size() - 1; i >= 0; --i) {
    cout << b[i];
  }
  cout << endl;
}
