#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, s;
  cin >> m >> s;
  if (s == 0 && m == 1) {
    cout << "0 0" << endl;
    return 0;
  } else if (s > 9 * m || s == 0) {
    cout << "-1 -1" << endl;
    return 0;
  }
  int totalsum = s;
  for (int i = m - 1; i >= 0; i--)
    for (int j = (i == m - 1 ? 1 : 0); j <= 9; j++)
      if (s - j <= 9 * i) {
        cout << j;
        s -= j;
        break;
      }
  s = totalsum;
  cout << " ";
  for (int i = m - 1; i >= 0; i--)
    for (int j = 9; j >= 0; j--)
      if (s - j >= 0) {
        cout << j;
        s -= j;
        break;
      }
  cout << endl;
}

### Prompt

Create a solution in cpp for the following problem:

You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the largest of the numbers that have length m and sum of digits s. The required numbers should be non-negative integers written in the decimal base without leading zeroes.

Input

The single line of the input contains a pair of integers m, s (1 ≤ m ≤ 100, 0 ≤ s ≤ 900) — the length and the sum of the digits of the required numbers.

Output

In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).

Examples

Input

2 15


Output

69 96


Input

3 0


Output

-1 -1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, s;
  cin >> m >> s;
  if (s == 0 && m == 1) {
    cout << "0 0" << endl;
    return 0;
  } else if (s > 9 * m || s == 0) {
    cout << "-1 -1" << endl;
    return 0;
  }
  int totalsum = s;
  for (int i = m - 1; i >= 0; i--)
    for (int j = (i == m - 1 ? 1 : 0); j <= 9; j++)
      if (s - j <= 9 * i) {
        cout << j;
        s -= j;
        break;
      }
  s = totalsum;
  cout << " ";
  for (int i = m - 1; i >= 0; i--)
    for (int j = 9; j >= 0; j--)
      if (s - j >= 0) {
        cout << j;
        s -= j;
        break;
      }
  cout << endl;
}

```