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;
int gcd(int x, int y) {
  if (y == 0) {
    return x;
  }
  return gcd(y, x % y);
}
long long lgcd(long long x, long long y) {
  if (y == 0) {
    return x;
  }
  return lgcd(y, x % y);
}
long long pow(long long a, long long b, long long m = 1000000007) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % m;
    }
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long modInverse(long long a, long long m = 1000000007) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  long long res = (x % m + m) % m;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, n, m;
  cin >> n >> m;
  string a = "", b = "";
  for (i = 1; i < n; i++) a += '8';
  a += '9';
  for (i = 0; i < n; i++) b += '1';
  cout << a << "\n" << b;
  return 0;
}
