Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int mod;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int mult(long long int p, long long int n) {
  long long int res = 1;
  p = (p % mod);
  while (n) {
    if (n % 2 == 1) res = (res * p) % (mod);
    p = (p * p) % (mod);
    n /= 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(7);
  int n, m;
  cin >> n >> m;
  string s1, s2;
  int num4 = (n / 4);
  for (int i = 0; i <= num4; i++) s1 += '4';
  s1 += '5';
  for (int i = 0; i < s1.length(); i++) s2 += '5';
  cout << s1 << " " << s2 << "\n";
  return 0;
}