#include <bits/stdc++.h>
using namespace std;
void F(int k, int n) {
  int r = 9;
  if (k == 1 && n == 0) {
    cout << 0;
  } else if (k * r >= n && k != 0 && n != 0) {
    while (k) {
      if (r > n) {
        r = n;
      }
      cout << r;
      n -= r;
      k--;
    }
  } else
    cout << -1;
}
void F2(int k, int n) {
  int r = 9, A[k + 1], q = k;
  if (n == 0 && k == 1)
    cout << 0;
  else if (k * r >= n && n != 0 && k != 0) {
    while (n) {
      if (r > n) {
        r = n;
      }
      A[k] = r;
      n -= r;
      k--;
    }
    if (k == 0) {
      for (int x = 1; x <= q; x++) {
        cout << A[x];
      }
    } else {
      cout << 1;
      for (int i = 1; i < k; i++) {
        cout << 0;
      }
      cout << A[k + 1] - 1;
      for (int i = k + 2; i <= q; i++) {
        cout << A[i];
      }
    }
  } else {
    cout << -1;
  }
  cout << " ";
}
int main() {
  int k, n;
  cin >> k >> n;
  F2(k, n);
  F(k, n);
}

### Prompt

Create a solution in Cpp for 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 F(int k, int n) {
  int r = 9;
  if (k == 1 && n == 0) {
    cout << 0;
  } else if (k * r >= n && k != 0 && n != 0) {
    while (k) {
      if (r > n) {
        r = n;
      }
      cout << r;
      n -= r;
      k--;
    }
  } else
    cout << -1;
}
void F2(int k, int n) {
  int r = 9, A[k + 1], q = k;
  if (n == 0 && k == 1)
    cout << 0;
  else if (k * r >= n && n != 0 && k != 0) {
    while (n) {
      if (r > n) {
        r = n;
      }
      A[k] = r;
      n -= r;
      k--;
    }
    if (k == 0) {
      for (int x = 1; x <= q; x++) {
        cout << A[x];
      }
    } else {
      cout << 1;
      for (int i = 1; i < k; i++) {
        cout << 0;
      }
      cout << A[k + 1] - 1;
      for (int i = k + 2; i <= q; i++) {
        cout << A[i];
      }
    }
  } else {
    cout << -1;
  }
  cout << " ";
}
int main() {
  int k, n;
  cin >> k >> n;
  F2(k, n);
  F(k, n);
}

```