#include <bits/stdc++.h>
using namespace std;
int num[105];
int main() {
  int m, s;
  while (scanf("%d%d", &m, &s) != EOF) {
    if (m == 1 && s == 0) {
      cout << 0 << ' ' << 0 << endl;
      continue;
    } else if (m * 9 < s || s < 1) {
      cout << -1 << ' ' << -1 << endl;
      continue;
    }
    for (int i = m - 1; i >= 0; i--) {
      num[i] = min(s, 9);
      s -= min(s, 9);
    }
    int k;
    for (k = 0; !num[k]; k++)
      ;
    num[0]++;
    num[k]--;
    for (int i = 0; i < m; i++) {
      cout << num[i];
    }
    cout << ' ';
    num[0]--;
    num[k]++;
    for (int i = m - 1; i >= 0; i--) {
      cout << num[i];
    }
    cout << endl;
  }
  return 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 num[105];
int main() {
  int m, s;
  while (scanf("%d%d", &m, &s) != EOF) {
    if (m == 1 && s == 0) {
      cout << 0 << ' ' << 0 << endl;
      continue;
    } else if (m * 9 < s || s < 1) {
      cout << -1 << ' ' << -1 << endl;
      continue;
    }
    for (int i = m - 1; i >= 0; i--) {
      num[i] = min(s, 9);
      s -= min(s, 9);
    }
    int k;
    for (k = 0; !num[k]; k++)
      ;
    num[0]++;
    num[k]--;
    for (int i = 0; i < m; i++) {
      cout << num[i];
    }
    cout << ' ';
    num[0]--;
    num[k]++;
    for (int i = m - 1; i >= 0; i--) {
      cout << num[i];
    }
    cout << endl;
  }
  return 0;
}

```