#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, m;
  cin >> n >> m;
  if (m == 0 && n == 1) {
    cout << "0 0";
  } else if (m > 9 * n || m == 0) {
    cout << "-1 -1";
    return;
  } else {
    long long a[n], b[n];
    long long sum = m, maxi = 9;
    for (long long i = 0; i < n; ++i) {
      a[i] = min(maxi, sum);
      sum -= a[i];
    }
    sum = m;
    for (int i = n - 1; i >= 1; --i) {
      b[i] = min(maxi, sum - 1);
      sum -= b[i];
    }
    b[0] = sum;
    for (int i = 0; i < n; ++i) {
      cout << b[i];
    }
    cout << " ";
    for (int i = 0; i < n; ++i) {
      cout << a[i];
    }
    cout << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  { solve(); }
  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;
void solve() {
  long long n, m;
  cin >> n >> m;
  if (m == 0 && n == 1) {
    cout << "0 0";
  } else if (m > 9 * n || m == 0) {
    cout << "-1 -1";
    return;
  } else {
    long long a[n], b[n];
    long long sum = m, maxi = 9;
    for (long long i = 0; i < n; ++i) {
      a[i] = min(maxi, sum);
      sum -= a[i];
    }
    sum = m;
    for (int i = n - 1; i >= 1; --i) {
      b[i] = min(maxi, sum - 1);
      sum -= b[i];
    }
    b[0] = sum;
    for (int i = 0; i < n; ++i) {
      cout << b[i];
    }
    cout << " ";
    for (int i = 0; i < n; ++i) {
      cout << a[i];
    }
    cout << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  { solve(); }
  return 0;
}

```