#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, s;
  cin >> m >> s;
  int a[101], arr[101], i = 1, k = 0;
  int temp = s;
  if (m == 1 && s == 0) {
    cout << "0 0";
    return 0;
  }
  if (9 * m < s || s == 0) {
    cout << "-1 -1";
    return 0;
  }
  for (; i <= 9, k < m; i++) {
    if (9 * (m - (k + 1)) >= s - i) {
      arr[k] = i;
      k++;
      s -= i;
      i = -1;
    }
  }
  for (i = 9, k = 0; i >= 0, k < m; i--) {
    if (temp - i >= 0) {
      a[k] = i;
      k++;
      temp -= i;
      i = 10;
    }
  }
  for (i = 0; i < k; i++) cout << arr[i];
  cout << " ";
  for (i = 0; i < k; i++) cout << a[i];
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
  int a[101], arr[101], i = 1, k = 0;
  int temp = s;
  if (m == 1 && s == 0) {
    cout << "0 0";
    return 0;
  }
  if (9 * m < s || s == 0) {
    cout << "-1 -1";
    return 0;
  }
  for (; i <= 9, k < m; i++) {
    if (9 * (m - (k + 1)) >= s - i) {
      arr[k] = i;
      k++;
      s -= i;
      i = -1;
    }
  }
  for (i = 9, k = 0; i >= 0, k < m; i--) {
    if (temp - i >= 0) {
      a[k] = i;
      k++;
      temp -= i;
      i = 10;
    }
  }
  for (i = 0; i < k; i++) cout << arr[i];
  cout << " ";
  for (i = 0; i < k; i++) cout << a[i];
  return 0;
}

```