#include <bits/stdc++.h>
using namespace std;
int res[110], m, s;
int main() {
  cin >> m >> s;
  if (!s) {
    if (m == 1)
      cout << "0 0" << endl;
    else
      cout << "-1 -1" << endl;
    return 0;
  }
  if (s > m * 9) {
    cout << "-1 -1" << endl;
    return 0;
  }
  for (int i = 1; i <= 9; i++) {
    res[0] = i;
    int ts = s - i;
    for (int j = m - 1; j && ts; j--) {
      if (ts > 9)
        res[j] = 9;
      else
        res[j] = ts;
      ts -= res[j];
    }
    if (!ts) {
      for (int j = 0; j < m; j++) cout << res[j];
      cout << " ";
      break;
    }
  }
  int ts = s;
  memset(res, 0, sizeof res);
  for (int i = 0; i < m; i++) {
    if (ts) {
      if (ts > 9) {
        cout << 9;
        ts -= 9;
      } else {
        cout << ts;
        ts -= ts;
      }
    } else
      cout << 0;
  }
  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 res[110], m, s;
int main() {
  cin >> m >> s;
  if (!s) {
    if (m == 1)
      cout << "0 0" << endl;
    else
      cout << "-1 -1" << endl;
    return 0;
  }
  if (s > m * 9) {
    cout << "-1 -1" << endl;
    return 0;
  }
  for (int i = 1; i <= 9; i++) {
    res[0] = i;
    int ts = s - i;
    for (int j = m - 1; j && ts; j--) {
      if (ts > 9)
        res[j] = 9;
      else
        res[j] = ts;
      ts -= res[j];
    }
    if (!ts) {
      for (int j = 0; j < m; j++) cout << res[j];
      cout << " ";
      break;
    }
  }
  int ts = s;
  memset(res, 0, sizeof res);
  for (int i = 0; i < m; i++) {
    if (ts) {
      if (ts > 9) {
        cout << 9;
        ts -= 9;
      } else {
        cout << ts;
        ts -= ts;
      }
    } else
      cout << 0;
  }
  cout << endl;
  return 0;
}

```