#include <bits/stdc++.h>
using namespace std;
int power10(int n) {
  if (n == 0)
    return 1;
  else
    return (power10(n - 1) * 10);
}
int main() {
  long m, s, max, min, cntlarge, cntsmall;
  cin >> m >> s;
  if (m == 1 && s == 0)
    cout << "0 0\n";
  else if ((s < 1) || (s > 9 * m)) {
    cout << "-1 -1\n";
  } else {
    if ((s - 1) / 9 == m - 1) {
      cout << (s % 9 == 0 ? 9 : s % 9);
      for (int i = 0; i < m - 1; i++) cout << 9;
    } else {
      cout << 1;
      for (int i = 0; i < m - (s - 1) / 9 - 2; i++) cout << 0;
      cout << (s - 1) % 9;
      for (int i = 0; i < (s - 1) / 9; i++) cout << 9;
    }
    cout << " ";
    for (int i = 0; i < s / 9; i++) cout << 9;
    if (s / 9 < m) cout << s % 9;
    for (int i = 0; i < m - s / 9 - 1; i++) cout << 0;
    cout << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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 power10(int n) {
  if (n == 0)
    return 1;
  else
    return (power10(n - 1) * 10);
}
int main() {
  long m, s, max, min, cntlarge, cntsmall;
  cin >> m >> s;
  if (m == 1 && s == 0)
    cout << "0 0\n";
  else if ((s < 1) || (s > 9 * m)) {
    cout << "-1 -1\n";
  } else {
    if ((s - 1) / 9 == m - 1) {
      cout << (s % 9 == 0 ? 9 : s % 9);
      for (int i = 0; i < m - 1; i++) cout << 9;
    } else {
      cout << 1;
      for (int i = 0; i < m - (s - 1) / 9 - 2; i++) cout << 0;
      cout << (s - 1) % 9;
      for (int i = 0; i < (s - 1) / 9; i++) cout << 9;
    }
    cout << " ";
    for (int i = 0; i < s / 9; i++) cout << 9;
    if (s / 9 < m) cout << s % 9;
    for (int i = 0; i < m - s / 9 - 1; i++) cout << 0;
    cout << endl;
  }
  return 0;
}

```