#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  int n;
  cin >> t >> n;
  if (t * 9 < n || (n == 0 && t > 1)) {
    cout << "-1 -1\n";
    return 0;
  }
  if (n == 0 && t == 1) {
    cout << "0 0\n";
    return 0;
  }
  string max = "";
  string ans = "";
  int k;
  int a = 9;
  for (int i = 0; i < t; i++) {
    k = min(n, a);
    n -= k;
    ans += ('0' + k);
  }
  max = ans;
  reverse(ans.begin(), ans.end());
  if (ans[0] == '0') {
    ans[0] = '1';
    for (int i = 1; i < t; i++) {
      if (ans[i] != '0') {
        ans[i] = (ans[i] - 1);
        break;
      }
    }
  }
  cout << ans << " " << max << endl;
  return 0;
}

### 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 t;
  int n;
  cin >> t >> n;
  if (t * 9 < n || (n == 0 && t > 1)) {
    cout << "-1 -1\n";
    return 0;
  }
  if (n == 0 && t == 1) {
    cout << "0 0\n";
    return 0;
  }
  string max = "";
  string ans = "";
  int k;
  int a = 9;
  for (int i = 0; i < t; i++) {
    k = min(n, a);
    n -= k;
    ans += ('0' + k);
  }
  max = ans;
  reverse(ans.begin(), ans.end());
  if (ans[0] == '0') {
    ans[0] = '1';
    for (int i = 1; i < t; i++) {
      if (ans[i] != '0') {
        ans[i] = (ans[i] - 1);
        break;
      }
    }
  }
  cout << ans << " " << max << endl;
  return 0;
}

```