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;
  long long size = 2 * n / 9 + 1;
  if (size < 3) {
    size = 3;
  }
  long long a[size];
  long long b[size];
  for (long long i = 0; i < size - 1; i++) {
    if (i & 1) {
      a[i] = 4;
      b[i] = 5;
    } else {
      a[i] = 5;
      b[i] = 4;
    }
  }
  if (n % 9 == 0) {
    a[size - 1] = 1;
    b[size - 1] = 9;
  } else {
    a[size - 1] = (n % 9);
    b[size - 1] = (10 - (n % 9));
  }
  for (long long i = 0; i < size; i++) {
    cout << a[i];
  }
  cout << "\n";
  for (long long i = 0; i < size; i++) {
    cout << b[i];
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}
