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;
const int MAXN = 1e8;
const int INF = 1e9;
int N, M;
int ss(int x) {
  int ans = 0;
  while (x) {
    ans += x - (x / 10) * 10;
    x /= 10;
  }
  return ans;
}
int cantdig(int a) {
  int digits = 0;
  while (a) {
    a /= 10;
    digits++;
  }
  return digits;
}
string S, E;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M;
  S.push_back('8');
  int digs = N;
  for (int i = digs - 1; i > 0; i--) {
    S.push_back('8');
  }
  S[S.size() - 1] += 1;
  for (int i = digs; i > 0; i--) {
    E.push_back('1');
  }
  cout << S << endl << E << endl;
  return 0;
}
