#include <bits/stdc++.h>
using namespace std;
int dp1[107], dp2[107];
int main() {
  int m, s;
  cin >> m >> s;
  if (m == 1 && s == 0) {
    cout << 0 << " " << 0 << endl;
    return 0;
  }
  bool check1 = 0, check2 = 0;
  if (s == 0) check2 = check1 = 1;
  int i = 0;
  int temp = 0;
  while (i < m) {
    if (temp == s) break;
    dp2[i]++;
    temp++;
    if (dp2[i] == 9) i++;
  }
  if (temp != s) check2 = 1;
  dp1[0] = 1;
  temp = 1;
  i = m - 1;
  while (i >= 0) {
    if (temp == s) break;
    dp1[i]++;
    temp++;
    if (dp1[i] == 9) i--;
  }
  if (temp != s) check1 = 1;
  if (check1)
    cout << -1;
  else
    for (int i = 0; i < m; i++) cout << dp1[i];
  cout << " ";
  if (check2)
    cout << -1;
  else
    for (int i = 0; i < m; i++) cout << dp2[i];
  cout << endl;
  return 0;
}

### Prompt

Please formulate 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 dp1[107], dp2[107];
int main() {
  int m, s;
  cin >> m >> s;
  if (m == 1 && s == 0) {
    cout << 0 << " " << 0 << endl;
    return 0;
  }
  bool check1 = 0, check2 = 0;
  if (s == 0) check2 = check1 = 1;
  int i = 0;
  int temp = 0;
  while (i < m) {
    if (temp == s) break;
    dp2[i]++;
    temp++;
    if (dp2[i] == 9) i++;
  }
  if (temp != s) check2 = 1;
  dp1[0] = 1;
  temp = 1;
  i = m - 1;
  while (i >= 0) {
    if (temp == s) break;
    dp1[i]++;
    temp++;
    if (dp1[i] == 9) i--;
  }
  if (temp != s) check1 = 1;
  if (check1)
    cout << -1;
  else
    for (int i = 0; i < m; i++) cout << dp1[i];
  cout << " ";
  if (check2)
    cout << -1;
  else
    for (int i = 0; i < m; i++) cout << dp2[i];
  cout << endl;
  return 0;
}

```