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;
struct edge {
  long long int A;
  long long int B;
};
long long int power(long long int a, long long int b, long long int mod) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2) {
      x = (x * y) % mod;
    }
    y = (y * y) % mod;
    b /= 2;
  }
  return x % mod;
}
long long int modular_inverse(long long int n, long long int mod) {
  return power(n, mod - 2, mod);
}
bool edgecompare(edge lhs, edge rhs) {
  if (lhs.B != rhs.B)
    return lhs.B < rhs.B;
  else
    return lhs.A < rhs.A;
}
long long int sod(long long int n) {
  long long int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j, k, flag = 0, x;
  cin >> n >> m;
  long long int a = 0, b = 0, count = 1;
  while (b < (n - 5)) {
    count++;
    b += 4;
  }
  for (i = 0; i < count; i++) {
    cout << "5";
  }
  cout << endl;
  for (i = 0; i < count - 1; i++) cout << "4";
  cout << "5";
  return 0;
}