#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, sum;
  cin >> n >> sum;
  if (n == 1 && sum == 0) {
    cout << "0 0";
    return 0;
  }
  if (sum < 1 || sum > 9 * n) {
    cout << "-1 -1";
    return 0;
  }
  vector<int> arr(n, 0);
  vector<int> a(n, 0);
  arr[0] = 1;
  int add = sum - 1;
  int j = n - 1;
  while (add >= 9) {
    arr[j] = 9;
    add = add - 9;
    j--;
  }
  arr[j] = arr[j] + add;
  for (int i = 0; i < n; i++) cout << arr[i];
  cout << " ";
  int i = 0;
  while (sum >= 9) {
    a[i] = 9;
    sum = sum - 9;
    i++;
  }
  a[i] = sum;
  for (int i = 0; i < n; i++) cout << a[i];
}

### Prompt

Generate 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 n, sum;
  cin >> n >> sum;
  if (n == 1 && sum == 0) {
    cout << "0 0";
    return 0;
  }
  if (sum < 1 || sum > 9 * n) {
    cout << "-1 -1";
    return 0;
  }
  vector<int> arr(n, 0);
  vector<int> a(n, 0);
  arr[0] = 1;
  int add = sum - 1;
  int j = n - 1;
  while (add >= 9) {
    arr[j] = 9;
    add = add - 9;
    j--;
  }
  arr[j] = arr[j] + add;
  for (int i = 0; i < n; i++) cout << arr[i];
  cout << " ";
  int i = 0;
  while (sum >= 9) {
    a[i] = 9;
    sum = sum - 9;
    i++;
  }
  a[i] = sum;
  for (int i = 0; i < n; i++) cout << a[i];
}

```