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;
const int INF = 1e9 + 5;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
void solve() {
  long long n, m;
  cin >> n >> m;
  string a, b;
  a.push_back('5');
  b.push_back('5');
  long long i = 0;
  long long sa = 5, sb = 5;
  while (sa < n || sb < n) {
    if (i % 2 == 0) {
      a.push_back('4');
      b.push_back('5');
      sa += 4;
      sb += 5;
    } else {
      a.push_back('5');
      b.push_back('4');
      sa += 5;
      sb += 4;
    }
    i++;
  }
  reverse(a.begin(), a.end());
  reverse(b.begin(), b.end());
  cout << a << "\n" << b << "\n";
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
