#include <bits/stdc++.h>
using namespace std;
char mini[110], maxi[110];
int main() {
  int m, s, curr;
  scanf("%d%d", &m, &s);
  if (m == 1 && s == 0) {
    printf("0 0");
    return 0;
  }
  if (s == 0 || s > (9 * m)) {
    printf("-1 -1");
    return 0;
  }
  curr = 0;
  for (int i = 0; i < m; i++) {
    mini[i] = max(s - curr - ((m - 1 - i) * 9), i == 0 ? 1 : 0) + '0';
    curr += mini[i] - '0';
  }
  mini[m] = '\0';
  curr = 0;
  for (int i = 0; i < m; i++) {
    maxi[i] = min(s - curr - ((m - 1 - i) * 0), 9) + '0';
    curr += maxi[i] - '0';
  }
  maxi[m] = '\0';
  printf("%s %s", mini, maxi);
  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;
char mini[110], maxi[110];
int main() {
  int m, s, curr;
  scanf("%d%d", &m, &s);
  if (m == 1 && s == 0) {
    printf("0 0");
    return 0;
  }
  if (s == 0 || s > (9 * m)) {
    printf("-1 -1");
    return 0;
  }
  curr = 0;
  for (int i = 0; i < m; i++) {
    mini[i] = max(s - curr - ((m - 1 - i) * 9), i == 0 ? 1 : 0) + '0';
    curr += mini[i] - '0';
  }
  mini[m] = '\0';
  curr = 0;
  for (int i = 0; i < m; i++) {
    maxi[i] = min(s - curr - ((m - 1 - i) * 0), 9) + '0';
    curr += maxi[i] - '0';
  }
  maxi[m] = '\0';
  printf("%s %s", mini, maxi);
  return 0;
}

```