#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(0);
  int m, s;
  cin >> m >> s;
  int z = m;
  int y = s;
  if ((m * 9 < s) || (m > 1 && s == 0))
    cout << "-1 -1";
  else {
    if (s < (m - 1) * 9) {
      cout << 1;
      s--;
    } else {
      int k = s - 9 * (m - 1);
      cout << k;
      s -= k;
    }
    m--;
    while (m) {
      if (9 * (m - 1) > s)
        cout << 0;
      else {
        int k = s - 9 * (m - 1);
        cout << k;
        s -= k;
      }
      m--;
    }
    cout << " ";
    while (z) {
      cout << max(0, min(9, y));
      y -= 9;
      z--;
    }
  }
}

### Prompt

Your task is to create 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 gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(0);
  int m, s;
  cin >> m >> s;
  int z = m;
  int y = s;
  if ((m * 9 < s) || (m > 1 && s == 0))
    cout << "-1 -1";
  else {
    if (s < (m - 1) * 9) {
      cout << 1;
      s--;
    } else {
      int k = s - 9 * (m - 1);
      cout << k;
      s -= k;
    }
    m--;
    while (m) {
      if (9 * (m - 1) > s)
        cout << 0;
      else {
        int k = s - 9 * (m - 1);
        cout << k;
        s -= k;
      }
      m--;
    }
    cout << " ";
    while (z) {
      cout << max(0, min(9, y));
      y -= 9;
      z--;
    }
  }
}

```