#include <bits/stdc++.h>
using namespace std;
int c = 0, k = 0;
int a[100] = {0};
void large(int num) {
  if (c == 0) return;
  if (num <= 9)
    a[k++] = num;
  else
    a[k++] = 9;
  c--;
  large(num - a[k - 1]);
  return;
}
int main() {
  int m, s;
  cin >> m >> s;
  if (s == 0 && m == 1)
    cout << 0 << " " << 0;
  else if (s < 1 || s > m * 9)
    cout << -1 << " " << -1;
  else {
    for (int i = 1; i < 10; i++) {
      c = m - 1;
      k = 1;
      a[0] = i;
      large(s - i);
      int sum = 0;
      for (int j = 0; j < k; j++) sum = sum + a[j];
      if (sum == s) break;
    }
    cout << a[0];
    for (int i = k - 1; i >= 1; i--) cout << a[i];
    cout << " ";
    c = m;
    k = 0;
    large(s);
    for (int i = 0; i < k; i++) cout << a[i];
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 c = 0, k = 0;
int a[100] = {0};
void large(int num) {
  if (c == 0) return;
  if (num <= 9)
    a[k++] = num;
  else
    a[k++] = 9;
  c--;
  large(num - a[k - 1]);
  return;
}
int main() {
  int m, s;
  cin >> m >> s;
  if (s == 0 && m == 1)
    cout << 0 << " " << 0;
  else if (s < 1 || s > m * 9)
    cout << -1 << " " << -1;
  else {
    for (int i = 1; i < 10; i++) {
      c = m - 1;
      k = 1;
      a[0] = i;
      large(s - i);
      int sum = 0;
      for (int j = 0; j < k; j++) sum = sum + a[j];
      if (sum == s) break;
    }
    cout << a[0];
    for (int i = k - 1; i >= 1; i--) cout << a[i];
    cout << " ";
    c = m;
    k = 0;
    large(s);
    for (int i = 0; i < k; i++) cout << a[i];
  }
  return 0;
}

```