#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<int> v;
  int a[200], sum = 0, n, m, coun = 0, i, j;
  cin >> n >> m;
  if (n * 9 < m || m < 1 && n > 1) {
    cout << -1 << " " << -1 << endl;
    return 0;
  }
  memset(a, 0, sizeof(a));
  a[0] = 1;
  m--;
  for (i = n - 1; i >= 0; i--) {
    if (m >= 9)
      a[i] = 9, m -= 9;
    else {
      a[i] += m;
      break;
    }
  }
  for (i = 0; i < n; i++) cout << a[i];
  cout << " ";
  for (i = n - 1; i >= 0; i--) {
    if (a[i] == 0) {
      coun++;
      continue;
    }
    v.push_back(a[i]);
  }
  if (v.size() == 0) {
    cout << 0;
    return 0;
  }
  for (i = 0; i < v.size() - 1; i++) {
    for (j = v.size() - 1; j > i; j--) {
      if (v[i] == 9) break;
      if (v[j] == 0) continue;
      v[i] += v[j];
      if (v[i] >= 9) {
        v[j] = v[i] - 9;
        v[i] = 9;
      } else
        v[j] = 0;
    }
  }
  for (i = 0; i < v.size(); i++) cout << v[i];
  while (coun--) cout << 0;
  return 0;
}

### Prompt

Your challenge is to write 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 main() {
  vector<int> v;
  int a[200], sum = 0, n, m, coun = 0, i, j;
  cin >> n >> m;
  if (n * 9 < m || m < 1 && n > 1) {
    cout << -1 << " " << -1 << endl;
    return 0;
  }
  memset(a, 0, sizeof(a));
  a[0] = 1;
  m--;
  for (i = n - 1; i >= 0; i--) {
    if (m >= 9)
      a[i] = 9, m -= 9;
    else {
      a[i] += m;
      break;
    }
  }
  for (i = 0; i < n; i++) cout << a[i];
  cout << " ";
  for (i = n - 1; i >= 0; i--) {
    if (a[i] == 0) {
      coun++;
      continue;
    }
    v.push_back(a[i]);
  }
  if (v.size() == 0) {
    cout << 0;
    return 0;
  }
  for (i = 0; i < v.size() - 1; i++) {
    for (j = v.size() - 1; j > i; j--) {
      if (v[i] == 9) break;
      if (v[j] == 0) continue;
      v[i] += v[j];
      if (v[i] >= 9) {
        v[j] = v[i] - 9;
        v[i] = 9;
      } else
        v[j] = 0;
    }
  }
  for (i = 0; i < v.size(); i++) cout << v[i];
  while (coun--) cout << 0;
  return 0;
}

```