#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, n, i, j, k, a[1000];
  cin >> m >> n;
  memset(a, 0, sizeof(a));
  if (m == 1 and n == 0) {
    cout << "0 0" << endl;
    return 0;
  }
  if (n == 0 || m * 9 < n)
    cout << -1 << " " << -1 << endl;
  else {
    j = n / 9;
    for (i = 0; i < j; i++) {
      a[i] = 9;
    }
    k = n % 9;
    a[j] = k;
    if (k == 0) {
      a[j - 1] -= 1;
    } else
      a[j] -= 1;
    a[m - 1] += 1;
    for (i = m - 1; i >= 0; i--) {
      cout << a[i];
    }
    cout << " ";
    if (k == 0)
      a[j - 1] += 1;
    else
      a[j] += 1;
    a[m - 1] -= 1;
    for (i = 0; i < m; i++) {
      cout << a[i];
    }
    cout << endl;
  }
  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, n, i, j, k, a[1000];
  cin >> m >> n;
  memset(a, 0, sizeof(a));
  if (m == 1 and n == 0) {
    cout << "0 0" << endl;
    return 0;
  }
  if (n == 0 || m * 9 < n)
    cout << -1 << " " << -1 << endl;
  else {
    j = n / 9;
    for (i = 0; i < j; i++) {
      a[i] = 9;
    }
    k = n % 9;
    a[j] = k;
    if (k == 0) {
      a[j - 1] -= 1;
    } else
      a[j] -= 1;
    a[m - 1] += 1;
    for (i = m - 1; i >= 0; i--) {
      cout << a[i];
    }
    cout << " ";
    if (k == 0)
      a[j - 1] += 1;
    else
      a[j] += 1;
    a[m - 1] -= 1;
    for (i = 0; i < m; i++) {
      cout << a[i];
    }
    cout << endl;
  }
  return 0;
}

```