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;
int main() {
  int n, m, i, j, ans;
  cin >> n >> m;
  vector<int> a, b;
  int tmp = n;
  while (tmp) {
    if (tmp >= 9) {
      a.push_back(9);
      tmp -= 9;
    } else {
      a.push_back(tmp);
      tmp = 0;
    }
  }
  int val = 10 - a[0];
  b.push_back(val);
  int sum = val;
  for (i = 1; i < a.size(); i++) {
    b.push_back(10 - (a[i] + 1));
    sum += b[b.size() - 1];
  }
  if (sum < n) {
    while (sum < n) {
      b.push_back(9);
      sum += 9;
    }
  }
  reverse(a.begin(), a.end());
  reverse(b.begin(), b.end());
  for (i = 0; i < a.size(); i++) cout << a[i];
  cout << " ";
  for (i = 0; i < b.size(); i++) cout << b[i];
  return 0;
}
