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

### Prompt

In Cpp, your task is to solve 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, i, k, j, mn = 0, mx = 0;
  cin >> m >> s;
  if (m == 1 && s == 0) {
    cout << "0 0";
    return 0;
  }
  if (s > 9 * m || s < 1) {
    cout << "-1 -1";
    return 0;
  }
  for (i = 1; i <= m; i++) {
    k = 0;
    if (i == 1) {
      k = 1;
    }
    for (j = k; j < 10; j++) {
      if (s - mn - j <= 9 * (m - i)) {
        mn += j;
        cout << j;
        break;
      }
    }
  }
  cout << " ";
  for (i = 1; i <= m; i++) {
    for (j = 9; j >= 0; j--) {
      if (s - mx - j >= 0) {
        mx += j;
        cout << j;
        break;
      }
    }
  }
  return 0;
}

```