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 main() {
  int n, m;
  cin >> n >> m;
  int sn = n;
  int fir = 0;
  vector<int> ans;
  while (n > 0) {
    if (fir == 0) {
      ans.push_back(6);
      fir = 1;
      n -= 4;
    } else if (fir == 1) {
      ans.push_back(5);
      n -= 4;
    }
  }
  for (int i = ans.size() - 1; i >= 0; i--) {
    printf("%d", ans[i]);
  }
  printf("\n");
  fir = 0;
  int j = 0;
  vector<int> ans2;
  while (j < ans.size()) {
    ans2.push_back(4);
    j++;
  }
  for (int i = ans2.size() - 1; i >= 0; i--) {
    printf("%d", ans2[i]);
  }
  printf("\n");
}