#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int N, m;
  cin >> N >> m;
  int ar[N];
  if (N == 1 && m == 0) {
    cout << "0 0" << endl;
    return 0;
  }
  if (m <= 9 * N && m >= 1) {
    int m1 = m;
    int c = 9;
    for (int i = N - 1; i > 0; i--) {
      if (m > c) {
        ar[i] = c;
        m -= c;
      } else {
        c--;
        i++;
      }
    }
    ar[0] = m;
    for (int i = 0; i < N; i++) {
      cout << ar[i];
    }
    cout << " ";
    c = 9;
    for (int i = 0; i < N - 1; i++) {
      if (m1 >= c) {
        ar[i] = c;
        m1 -= c;
      } else {
        c--;
        i--;
      }
    }
    ar[N - 1] = m1;
    for (int i = 0; i < N; i++) {
      cout << ar[i];
    }
    cout << endl;
  } else {
    cout << "-1 -1" << endl;
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int N, m;
  cin >> N >> m;
  int ar[N];
  if (N == 1 && m == 0) {
    cout << "0 0" << endl;
    return 0;
  }
  if (m <= 9 * N && m >= 1) {
    int m1 = m;
    int c = 9;
    for (int i = N - 1; i > 0; i--) {
      if (m > c) {
        ar[i] = c;
        m -= c;
      } else {
        c--;
        i++;
      }
    }
    ar[0] = m;
    for (int i = 0; i < N; i++) {
      cout << ar[i];
    }
    cout << " ";
    c = 9;
    for (int i = 0; i < N - 1; i++) {
      if (m1 >= c) {
        ar[i] = c;
        m1 -= c;
      } else {
        c--;
        i--;
      }
    }
    ar[N - 1] = m1;
    for (int i = 0; i < N; i++) {
      cout << ar[i];
    }
    cout << endl;
  } else {
    cout << "-1 -1" << endl;
  }
}

```