#include <bits/stdc++.h>
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  int max[n], min[n];
  if (k == 0 || k > n * 9) {
    if (k == 0 && n == 1) {
      printf("0 0");
      return 0;
    }
    printf("-1 -1");
    return 0;
  }
  int k1 = k, i = 0;
  while (i < n) {
    if (k1 - 9 >= 0) {
      max[i] = 9;
      min[n - i - 1] = 9;
      i++;
      k1 -= 9;
    } else if (k1 == 0) {
      max[i] = 0;
      min[n - i - 1] = 0;
      i++;
    } else {
      max[i] = k1;
      min[n - i - 1] = k1;
      k1 = 0;
      i++;
    }
  }
  if (min[0] == 0) {
    min[0] = 1;
    i = 1;
    while (min[i] == 0) i++;
    min[i]--;
  }
  for (i = 0; i < n; i++) printf("%d", min[i]);
  printf(" ");
  for (i = 0; i < n; i++) printf("%d", max[i]);
}

### 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>
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  int max[n], min[n];
  if (k == 0 || k > n * 9) {
    if (k == 0 && n == 1) {
      printf("0 0");
      return 0;
    }
    printf("-1 -1");
    return 0;
  }
  int k1 = k, i = 0;
  while (i < n) {
    if (k1 - 9 >= 0) {
      max[i] = 9;
      min[n - i - 1] = 9;
      i++;
      k1 -= 9;
    } else if (k1 == 0) {
      max[i] = 0;
      min[n - i - 1] = 0;
      i++;
    } else {
      max[i] = k1;
      min[n - i - 1] = k1;
      k1 = 0;
      i++;
    }
  }
  if (min[0] == 0) {
    min[0] = 1;
    i = 1;
    while (min[i] == 0) i++;
    min[i]--;
  }
  for (i = 0; i < n; i++) printf("%d", min[i]);
  printf(" ");
  for (i = 0; i < n; i++) printf("%d", max[i]);
}

```