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 n, m;
vector<int> x, y;
int main() {
  scanf("%d%d", &n, &m);
  int s = 0, s2 = 0;
  int p = 0;
  for (;;) {
    if (p % 2 == 0) {
      s += 8;
      s2++;
      x.push_back(8);
      y.push_back(1);
    } else {
      s++;
      s2 += 8;
      x.push_back(1);
      y.push_back(8);
    }
    if (s >= n - 1 && s2 >= n) break;
    p ^= 1;
  }
  x[0]++;
  reverse(x.begin(), x.end());
  reverse(y.begin(), y.end());
  for (int o : x) printf("%d", o);
  printf("\n");
  for (int o : y) printf("%d", o);
  printf("\n");
  return 0;
}
