#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, s, i, s2;
  scanf("%d %d", &m, &s);
  s2 = s;
  if (s > m * 9 || (s == 0 && m > 1))
    printf("-1 -1");
  else {
    if (s < (m - 1) * 9) {
      printf("1");
      s--;
    } else {
      printf("%d", s - (m - 1) * 9);
      s = (m - 1) * 9;
    }
    for (i = 2; i <= m; i++) {
      if (s <= (m - i) * 9)
        printf("0");
      else {
        printf("%d", s - (m - i) * 9);
        s = (m - i) * 9;
      }
    }
    printf(" ");
    for (i = 1; i <= m; i++) {
      if (s2 > 9) {
        printf("9");
        s2 = s2 - 9;
      } else {
        printf("%d", s2);
        s2 = 0;
      }
    }
  }
}

### Prompt

Please formulate 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, s2;
  scanf("%d %d", &m, &s);
  s2 = s;
  if (s > m * 9 || (s == 0 && m > 1))
    printf("-1 -1");
  else {
    if (s < (m - 1) * 9) {
      printf("1");
      s--;
    } else {
      printf("%d", s - (m - 1) * 9);
      s = (m - 1) * 9;
    }
    for (i = 2; i <= m; i++) {
      if (s <= (m - i) * 9)
        printf("0");
      else {
        printf("%d", s - (m - i) * 9);
        s = (m - i) * 9;
      }
    }
    printf(" ");
    for (i = 1; i <= m; i++) {
      if (s2 > 9) {
        printf("9");
        s2 = s2 - 9;
      } else {
        printf("%d", s2);
        s2 = 0;
      }
    }
  }
}

```