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;
inline long long int read() {
  char c = getchar();
  long long int num, sign = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') sign = -1;
  for (num = 0; c >= '0' && c <= '9';) {
    c -= '0';
    num = num * 10 + c;
    c = getchar();
  }
  return num * sign;
}
inline long long int pw(long long int b, long long int p) {
  long long int ans = 1;
  long long int i = 0;
  while (i < p) ans *= b, i++;
  return ans;
}
inline bool isPrime(long long int n) {
  if (n == 1) return false;
  if (n == 2) return true;
  if (!(n % 2)) return false;
  for (long long int i = 3; i <= sqrt(n); i += 2)
    if (!(n % i)) return false;
  return true;
}
inline long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n = read(), m = read();
  int len = 300;
  cout << string(len, '5') << "\n";
  cout << (string(len - 1, '4') + '5') << "\n";
  return 0;
}
