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;
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a, b;
  for (__typeof((n)) i = (0); i < (n); i++) a.push_back(1);
  for (__typeof((n)) i = (0); i < (n); i++) b.push_back(8);
  b[n - 1]++;
  for (__typeof(((int)a.size())) i = (0); i < ((int)a.size()); i++)
    cout << a[i];
  cout << "" << endl;
  for (__typeof(((int)b.size())) i = (0); i < ((int)b.size()); i++)
    cout << b[i];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}
