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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
inline ll nextLong() {
  ll n;
  scanf("%lld", &n);
  return n;
}
inline int nextInt() {
  int n;
  scanf("%d", &n);
  return n;
}
inline double nextDouble() {
  double n;
  scanf("%lf", &n);
  return n;
}
inline string nextLine() {
  static char ______________buffer[1000005];
  fgets(______________buffer, 1000000, stdin);
  string res = ______________buffer;
  if (*(res.end() - 1) == '\n') res.erase(res.end() - 1);
  return res;
}
inline char nextChar() {
  char n;
  scanf("%c", &n);
  return n;
}
inline string next() {
  static char ______________buffer[1000005];
  scanf("%s", ______________buffer);
  string res = ______________buffer;
  return res;
}
void debug(ll n) {}
void debug(int n) {}
void debug(double n) {}
void debug(string n) {}
void debug(char n) {}
void debug(bool n) {}
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  for (int i = 0; i < a; i++) printf("1");
  printf("\n");
  for (int i = 0; i < a - 1; i++) printf("8");
  printf("9");
  return 0;
}
