#include <bits/stdc++.h>
using namespace std;
using T = unsigned;
std::string mi(T l, T s) {
  std::vector<char> num(l, '0');
  num[0] = '1';
  --s;
  auto idx = num.size();
  while (s > 9) {
    num[--idx] = '9';
    s -= 9;
  }
  num[--idx] += s;
  return std::string(num.begin(), num.end());
}
std::string ma(T l, T s) {
  std::vector<char> num(l, '0');
  auto idx = 0;
  while (s > 9) {
    num[idx++] = '9';
    s -= 9;
  }
  num[idx] += s;
  return std::string(num.begin(), num.end());
}
int main() {
  T l, s;
  std::cin >> l >> s;
  if (s == 0 && l == 1) {
    std::cout << "0 0";
    return 0;
  }
  if (s < 1 || s > 9 * l) {
    std::cout << -1 << ' ' << -1;
    return 0;
  }
  std::cout << mi(l, s) << ' ' << ma(l, s);
}

### 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;
using T = unsigned;
std::string mi(T l, T s) {
  std::vector<char> num(l, '0');
  num[0] = '1';
  --s;
  auto idx = num.size();
  while (s > 9) {
    num[--idx] = '9';
    s -= 9;
  }
  num[--idx] += s;
  return std::string(num.begin(), num.end());
}
std::string ma(T l, T s) {
  std::vector<char> num(l, '0');
  auto idx = 0;
  while (s > 9) {
    num[idx++] = '9';
    s -= 9;
  }
  num[idx] += s;
  return std::string(num.begin(), num.end());
}
int main() {
  T l, s;
  std::cin >> l >> s;
  if (s == 0 && l == 1) {
    std::cout << "0 0";
    return 0;
  }
  if (s < 1 || s > 9 * l) {
    std::cout << -1 << ' ' << -1;
    return 0;
  }
  std::cout << mi(l, s) << ' ' << ma(l, s);
}

```