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;
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
bool isprime(long long p) {
  if (p < 2) return false;
  if (p == 2) return true;
  if (p % 2 == 0) return false;
  double limit = sqrt(p);
  for (long long i = 3; i <= limit; i += 2) {
    if ((p % i) == 0) return false;
  }
  return true;
}
long long fac(long long n) {
  long long an = 1;
  while (n > 1) {
    an *= n;
    n--;
  }
  return an;
}
long long kor(long long n, long long st) {
  long long res = 1;
  while (st) {
    if (st & 1) res *= n;
    n *= n;
    st >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  string ans1, ans2;
  for (int i = 0; i < 2200; ++i) {
    ans1 += "4";
    ans2 += "5";
  }
  ans1 = ans1 + "5";
  ans2 += "5";
  cout << ans1 << "\n" << ans2;
  return 0;
}
