#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, s, i;
  int temp;
  int *c;
  cin >> m >> s;
  temp = s;
  if (s > m * 9 || s < 0 || (s == 0 && m > 1))
    cout << "-1 -1";
  else {
    c = new int[m];
    for (i = m - 1; s > 9; i--) {
      c[i] = 9;
      s -= 9;
    }
    c[i] = s;
    if (i > 0) {
      c[i--]--;
      while (i > 0) c[i--] = 0;
      c[i] = 1;
    }
    i = 0;
    while (i < m) cout << c[i++];
    cout << ' ';
    i = 0;
    s = temp;
    while (s > 9) {
      c[i++] = 9;
      s -= 9;
    }
    c[i++] = s;
    while (i < m) c[i++] = 0;
    i = 0;
    while (i < m) cout << c[i++];
  }
  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 m, s, i;
  int temp;
  int *c;
  cin >> m >> s;
  temp = s;
  if (s > m * 9 || s < 0 || (s == 0 && m > 1))
    cout << "-1 -1";
  else {
    c = new int[m];
    for (i = m - 1; s > 9; i--) {
      c[i] = 9;
      s -= 9;
    }
    c[i] = s;
    if (i > 0) {
      c[i--]--;
      while (i > 0) c[i--] = 0;
      c[i] = 1;
    }
    i = 0;
    while (i < m) cout << c[i++];
    cout << ' ';
    i = 0;
    s = temp;
    while (s > 9) {
      c[i++] = 9;
      s -= 9;
    }
    c[i++] = s;
    while (i < m) c[i++] = 0;
    i = 0;
    while (i < m) cout << c[i++];
  }
  return 0;
}

```