#include <bits/stdc++.h>
using namespace std;
int n, s, a[11111];
string s1;
int main() {
  cin >> n >> s;
  if ((n > 1 && s == 0) || 9 * n < s) {
    cout << -1 << ' ' << -1;
    return 0;
  }
  int i = 0, ss = s;
  while (ss >= 9) {
    a[i] = 9;
    s1 += '9';
    ss -= 9;
    i++;
  }
  a[i] = ss;
  if (ss || i != n) {
    s1 += char(ss + '0');
    i++;
  }
  for (int j = i; j < n; j++) {
    s1 += '0';
    a[j] = 0;
  }
  sort(a, a + n);
  if (!a[0]) {
    int j = n - 1;
    while (a[j] > 0) {
      j--;
    }
    if (a[j + 1]) {
      a[0] = 1;
      a[j + 1]--;
    }
  }
  for (int i = 0; i < n; i++) cout << a[i];
  cout << ' ' << s1;
  cout << endl;
}

### 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 n, s, a[11111];
string s1;
int main() {
  cin >> n >> s;
  if ((n > 1 && s == 0) || 9 * n < s) {
    cout << -1 << ' ' << -1;
    return 0;
  }
  int i = 0, ss = s;
  while (ss >= 9) {
    a[i] = 9;
    s1 += '9';
    ss -= 9;
    i++;
  }
  a[i] = ss;
  if (ss || i != n) {
    s1 += char(ss + '0');
    i++;
  }
  for (int j = i; j < n; j++) {
    s1 += '0';
    a[j] = 0;
  }
  sort(a, a + n);
  if (!a[0]) {
    int j = n - 1;
    while (a[j] > 0) {
      j--;
    }
    if (a[j + 1]) {
      a[0] = 1;
      a[j + 1]--;
    }
  }
  for (int i = 0; i < n; i++) cout << a[i];
  cout << ' ' << s1;
  cout << endl;
}

```