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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int i = 0, k, q1, t, q, m, x, y, r, n, cnt = 0, cnt1 = 0, p,
                sum = 0, sum1 = 0, sum2 = 0, j, ans = 0, a, b, ans1;
  q1 = 1;
  while (q1--) {
    cin >> n >> m;
    ans = n / 4;
    ans1 = n % 4;
    for (int i = 0; i < ans; i++) {
      cout << 4;
    }
    if (ans1 != 0)
      cout << ans1 << " ";
    else
      cout << " ";
    for (int i = 0; i < ans; i++) {
      if (ans1 == 0 && i == ans - 1)
        cout << 6;
      else
        cout << 5;
    }
    if (ans1 != 0) cout << 10 - ans1;
  }
}
