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 M = 1e9 + 7;
int mul(int a, int b) { return (long long int)a * b % M; }
int power(int a, int b) {
  if (b == 0) return 1;
  int temp = power(a, b / 2);
  if (b % 2 == 0)
    return mul(temp, temp);
  else
    return mul(a, mul(temp, temp));
}
int inv(int v) { return power(v, M - 2); }
int main() {
  int a, b;
  cin >> a >> b;
  string s1, s2;
  s1 = s2 = "";
  for (int i = 0; i < a - 1; i++) {
    s2 += "1";
    s1 += "8";
  }
  s2 += "1";
  s1 += "9";
  cout << s1 << " " << s2 << endl;
}
