#include <bits/stdc++.h>
using namespace std;
void XsliferX() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
int dx[9] = {1, -1, 0, 0, 1, -1, 1, -1, 0};
int dy[9] = {0, 0, 1, -1, 1, -1, -1, 1, 0};
int mxarr[1000], mnarr[1000];
int main() {
  XsliferX();
  int n, s, ss;
  cin >> n >> s;
  if ((!s && n != 1) || n * 9 < s) return cout << "-1 -1", 0;
  ss = s;
  for (int i = 0; i < n; i++) {
    int temp = min(9, s);
    s -= temp;
    mxarr[i] += temp;
  }
  s = ss;
  s--;
  mnarr[n - 1] = 1;
  for (int i = 0; i < n; i++) {
    int temp = min(9, s);
    s -= temp;
    mnarr[i] += temp;
  }
  for (int i = n - 1; i >= 0; i--) {
    cout << mnarr[i];
  }
  cout << " ";
  for (int i = 0; i < n; i++) {
    cout << mxarr[i];
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 XsliferX() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
int dx[9] = {1, -1, 0, 0, 1, -1, 1, -1, 0};
int dy[9] = {0, 0, 1, -1, 1, -1, -1, 1, 0};
int mxarr[1000], mnarr[1000];
int main() {
  XsliferX();
  int n, s, ss;
  cin >> n >> s;
  if ((!s && n != 1) || n * 9 < s) return cout << "-1 -1", 0;
  ss = s;
  for (int i = 0; i < n; i++) {
    int temp = min(9, s);
    s -= temp;
    mxarr[i] += temp;
  }
  s = ss;
  s--;
  mnarr[n - 1] = 1;
  for (int i = 0; i < n; i++) {
    int temp = min(9, s);
    s -= temp;
    mnarr[i] += temp;
  }
  for (int i = n - 1; i >= 0; i--) {
    cout << mnarr[i];
  }
  cout << " ";
  for (int i = 0; i < n; i++) {
    cout << mxarr[i];
  }
  return 0;
}

```