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;
int n, m, tmp = 1;
int sum(int a) {
  int res = 0;
  while (a) {
    res += a % 10;
    a /= 10;
  }
  return res;
}
int main() {
  cin >> n >> m;
  if (2 * n > 2220) {
    int nn = n / 2 + n % 2;
    for (int i = 1; i <= nn; ++i) {
      cout << 2;
    }
    cout << endl;
    for (int i = 1; i <= nn - 1; ++i) {
      cout << 7;
    }
    cout << 8 << endl;
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    cout << 1;
  }
  cout << endl;
  for (int i = 1; i <= n - 1; ++i) {
    cout << 8;
  }
  cout << 9 << endl;
}