#include <bits/stdc++.h>
using namespace std;
const int MXN = 1000;
int mx[MXN];
int mn[MXN];
void find_mn(int m, int s) {
  mn[0] = 1;
  for (int i = 1; i < m; i++) {
    mn[i] = 0;
  }
  int sum = 1;
  int pt = m - 1;
  while (sum < s) {
    mn[pt]++;
    sum++;
    if (mn[pt] == 9) {
      pt--;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << mn[i];
  }
}
void find_mx(int m, int s) {
  for (int i = 0; i < m; i++) {
    mx[i] = 9;
  }
  int sum = 9 * m;
  int pt = m - 1;
  while (sum > s) {
    mx[pt]--;
    sum--;
    if (mx[pt] == 0) {
      pt--;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << mx[i];
  }
}
int main() {
  int m, s;
  cin >> m >> s;
  if (m == 1 && s == 0) {
    cout << "0 0\n";
    return 0;
  }
  if (s > 9 * m || s == 0) {
    cout << "-1 -1\n";
    return 0;
  }
  find_mn(m, s);
  cout << " ";
  find_mx(m, s);
  cout << "\n";
}

### 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;
const int MXN = 1000;
int mx[MXN];
int mn[MXN];
void find_mn(int m, int s) {
  mn[0] = 1;
  for (int i = 1; i < m; i++) {
    mn[i] = 0;
  }
  int sum = 1;
  int pt = m - 1;
  while (sum < s) {
    mn[pt]++;
    sum++;
    if (mn[pt] == 9) {
      pt--;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << mn[i];
  }
}
void find_mx(int m, int s) {
  for (int i = 0; i < m; i++) {
    mx[i] = 9;
  }
  int sum = 9 * m;
  int pt = m - 1;
  while (sum > s) {
    mx[pt]--;
    sum--;
    if (mx[pt] == 0) {
      pt--;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << mx[i];
  }
}
int main() {
  int m, s;
  cin >> m >> s;
  if (m == 1 && s == 0) {
    cout << "0 0\n";
    return 0;
  }
  if (s > 9 * m || s == 0) {
    cout << "-1 -1\n";
    return 0;
  }
  find_mn(m, s);
  cout << " ";
  find_mx(m, s);
  cout << "\n";
}

```