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;
inline int iin() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long int lin() {
  long long int x;
  scanf("%lld", &x);
  return x;
}
inline long long int Lin() {
  long long int x;
  scanf("%I64d", &x);
  return x;
}
inline int iout(int x) { printf("%d", x); }
inline long long int lout(long long int x) { printf("%lld", x); }
inline long long int Lout(long long int x) { printf("%I64d", x); }
const int maxn4 = 1e4 + 85, maxn5 = 1e5 + 69, maxn6 = 1e6 + 85,
          maxn7 = 1e7 + 69, maxn8 = 1e8 + 85, maxn9 = 1e9 + 69, modi1 = 1e7 + 7,
          modi2 = 1e9 + 7;
const long long int mod1 = 1e7 + 7, mod2 = 1e9 + 7;
int mn = INT_MAX, mx = INT_MIN;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cout << 1;
  cout << '\n';
  for (int i = 1; i < n; i++) cout << 8;
  cout << 9 << '\n';
  return 0;
}
