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;
const int MAX_N = 1e6 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const long long int LINF = 1e18;
const long long int N = 100001;
void snake_bite() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
}
bool containsDigit(long long int number, int digit) {
  while (number != 0) {
    long long int curr_digit = number % 10;
    if (curr_digit == digit) return true;
    number /= 10;
  }
  return false;
}
int sum_digit(int x) {
  int sum = 0;
  while (x) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cout << "1";
  cout << " ";
  for (int i = 0; i < n - 1; i++) cout << "8";
  cout << "9";
}
int main() {
  snake_bite();
  long int t = 1;
  while (t--) {
    solve();
    cout << endl;
  }
}