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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (int i = 0; i < v.size(); i++) os << v[i] << " ";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (typename set<T>::iterator it = v.begin(); it != v.end(); it++)
    os << *it << " ";
}
const double eps = 1e-9;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int main() {
  int n = gi(), m = gi();
  if (n > 1000) {
    n /= 2;
  }
  for (int i = (1); i <= (n); i++) {
    cout << "5";
  }
  cout << "\n";
  for (int i = (1); i <= (n - 1); i++) {
    cout << "4";
  }
  cout << "5\n";
  return 0;
}
