#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long m, s;
  cin >> m >> s;
  if (s == 0) {
    if (m == 1)
      cout << "0 0";
    else
      cout << "-1 -1";
    return 0;
  }
  string a(m, '0');
  a[0] = '1';
  long long cur = 1;
  for (long long i = m - 1; i >= 0; --i) {
    while (a[i] < '9' && cur < s) {
      a[i]++;
      cur++;
    }
  }
  if (cur < s) {
    cout << "-1 -1";
    return 0;
  }
  cout << a << ' ';
  string b(m, '0');
  cur = 0;
  for (long long i = 0; i < m; ++i) {
    while (b[i] < '9' && cur < s) {
      b[i]++;
      cur++;
    }
  }
  cout << b;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long m, s;
  cin >> m >> s;
  if (s == 0) {
    if (m == 1)
      cout << "0 0";
    else
      cout << "-1 -1";
    return 0;
  }
  string a(m, '0');
  a[0] = '1';
  long long cur = 1;
  for (long long i = m - 1; i >= 0; --i) {
    while (a[i] < '9' && cur < s) {
      a[i]++;
      cur++;
    }
  }
  if (cur < s) {
    cout << "-1 -1";
    return 0;
  }
  cout << a << ' ';
  string b(m, '0');
  cur = 0;
  for (long long i = 0; i < m; ++i) {
    while (b[i] < '9' && cur < s) {
      b[i]++;
      cur++;
    }
  }
  cout << b;
}

```