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;
const int N = 100010;
const int iinf = INT_MAX;
const long long linf = LONG_MAX;
const int MOD = 1e9 + 7;
inline int read() {
  int X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  if (n + n > 2230) {
    for (int i = 1; i <= n / 2 + 1; i++) cout << 2;
    cout << endl;
    for (int i = 1; i <= n / 2; i++) cout << 7;
    cout << 8 << endl;
  } else {
    for (int i = 1; i <= n; i++) cout << 1;
    cout << endl;
    for (int i = 1; i < n; i++) cout << 8;
    cout << 9 << endl;
  }
}
