#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, sum;
  cin >> m >> sum;
  if (sum == 0 && m > 1) {
    cout << "-1 -1" << endl;
    return 0;
  }
  if (sum > 9 * m) {
    cout << "-1 -1" << endl;
    return 0;
  }
  int s = sum;
  int res[m];
  s = s - 1;
  for (int i = m - 1; i > 0; i--) {
    if (s > 9) {
      res[i] = 9;
      s -= 9;
    } else {
      res[i] = s;
      s = 0;
    }
  }
  res[0] = s + 1;
  for (int i = 0; i < m; i++) cout << res[i];
  cout << " ";
  s = sum;
  for (int i = 0; i < m; i++) {
    if (s > 9) {
      res[i] = 9;
      s -= 9;
    } else {
      res[i] = s;
      s = 0;
    }
  }
  for (int i = 0; i < m; i++) cout << res[i];
  cout << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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, sum;
  cin >> m >> sum;
  if (sum == 0 && m > 1) {
    cout << "-1 -1" << endl;
    return 0;
  }
  if (sum > 9 * m) {
    cout << "-1 -1" << endl;
    return 0;
  }
  int s = sum;
  int res[m];
  s = s - 1;
  for (int i = m - 1; i > 0; i--) {
    if (s > 9) {
      res[i] = 9;
      s -= 9;
    } else {
      res[i] = s;
      s = 0;
    }
  }
  res[0] = s + 1;
  for (int i = 0; i < m; i++) cout << res[i];
  cout << " ";
  s = sum;
  for (int i = 0; i < m; i++) {
    if (s > 9) {
      res[i] = 9;
      s -= 9;
    } else {
      res[i] = s;
      s = 0;
    }
  }
  for (int i = 0; i < m; i++) cout << res[i];
  cout << endl;
  return 0;
}

```