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;
long long int div_floor(const long long int &a, const long long int &b) {
  return a / b - (((a ^ b) < 0) and a % b);
}
long long int div_ceil(const long long int &a, const long long int &b) {
  return a / b + (((a ^ b) >= 0) and a % b);
}
const long long int mod = 1e9 + 7;
void solving() {
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) cout << "1";
  cout << "\n";
  for (long long int i = 0; i < n - 1; i++) cout << "8";
  cout << "9\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solving();
  return 0;
}