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;
long long power(long long a, long long b, long long mod) {
  long long c = 1;
  while (b > 0) {
    if (b % 2) c *= a, c %= mod;
    b /= 2;
    a *= a;
    a %= mod;
  }
  return c;
}
const long long N = (5e3) + 10;
const long long M = 320;
const long long Mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(25);
  long long n, m, i, j, k, a, b, c, d;
  cin >> n >> m;
  string s, t;
  i = 300;
  j = 0;
  while (i--) {
    if (i % 2)
      s += '4', t += '5';
    else
      s += '5', t += '4';
    j += 4;
  }
  s += '5';
  t += '5';
  cout << s << ' ' << t << '\n';
  return 0;
}
