#include <bits/stdc++.h>
using namespace std;
long long m, s, x[1000], ok, flag;
int main() {
  cin >> m >> s;
  if (s == 0) {
    if (m == 1)
      cout << "0 0\n";
    else
      cout << "-1 -1\n";
    return 0;
  }
  for (int i = 0; i < m; i++) {
    x[i] = min((long long)9, s);
    s -= x[i];
    if (s == 0 && !ok) {
      ok++;
      flag = i;
    }
  }
  if (s) {
    cout << "-1 -1\n";
    return 0;
  }
  for (int j = m - 1; j >= 0; j--) {
    if (j == flag && x[m - 1] == 0)
      cout << x[j] - 1;
    else if (j == m - 1 && x[j] == 0)
      cout << 1;
    else
      cout << x[j];
  }
  cout << " ";
  for (int i = 0; i < m; i++) cout << x[i];
  cout << endl;
  return 0;
}

### 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;
long long m, s, x[1000], ok, flag;
int main() {
  cin >> m >> s;
  if (s == 0) {
    if (m == 1)
      cout << "0 0\n";
    else
      cout << "-1 -1\n";
    return 0;
  }
  for (int i = 0; i < m; i++) {
    x[i] = min((long long)9, s);
    s -= x[i];
    if (s == 0 && !ok) {
      ok++;
      flag = i;
    }
  }
  if (s) {
    cout << "-1 -1\n";
    return 0;
  }
  for (int j = m - 1; j >= 0; j--) {
    if (j == flag && x[m - 1] == 0)
      cout << x[j] - 1;
    else if (j == m - 1 && x[j] == 0)
      cout << 1;
    else
      cout << x[j];
  }
  cout << " ";
  for (int i = 0; i < m; i++) cout << x[i];
  cout << endl;
  return 0;
}

```