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;
void solve() {
  long long n, m;
  cin >> n >> m;
  long long x = 2 + ceil((n - 7) / 5.0);
  long long y = 2 + ceil((n - 3) / 4.0);
  if (x >= y) {
    string a = "7", b = "3";
    for (long long i = 1; i < x; i++) {
      a += "5";
      b += "4";
    }
    reverse(a.begin(), a.end());
    reverse(b.begin(), b.end());
    cout << a << "\n" << b << "\n";
  } else if (y > x) {
    string a = "7", b = "3";
    for (long long i = 1; i < y; i++) {
      a += "5";
      b += "4";
    }
    reverse(a.begin(), a.end());
    reverse(b.begin(), b.end());
    cout << a << "\n" << b << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}