#include <bits/stdc++.h>
using namespace std;
void small(long long m, long long s) {
  vector<long long> v;
  long long p = (s % 9 == 0) ? s / 9 - 1 : s / 9;
  for (long long i = 0; i < p; i++) {
    v.push_back(9);
    s -= 9;
  }
  if (m - p == 1) {
    v.push_back(s);
  }
  if (m - p >= 2) {
    s--;
    v.push_back(s);
    for (long long i = 0; i < m - p - 2; i++) {
      v.push_back(0);
    }
    v.push_back(1);
  }
  for (long long i = v.size() - 1; i >= 0; i--) {
    cout << v[i];
  }
  cout << " ";
}
void great(long long m, long long s) {
  vector<long long> v;
  long long k = s / 9;
  for (long long i = 0; i < k; i++) {
    v.push_back(9);
    s -= 9;
  }
  if (s > 0) {
    v.push_back(s);
    k++;
  }
  for (long long i = 0; i < m - k; i++) {
    v.push_back(0);
  }
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i];
  }
}
int main() {
  long long m, s, h;
  string k;
  vector<int> v;
  cin >> m >> s;
  if (s % 9 == 0) {
    long long t = 0;
    if (m == 1 && s == 0) {
      cout << "0 "
           << "0";
      t = 1;
    }
    if (t == 0) {
      if (m < s / 9 || s == 0) {
        cout << "-1 "
             << "-1";
      } else {
        small(m, s);
        great(m, s);
      }
    }
  } else if (s % 9 != 0) {
    long long t = 0;
    if (m == 1 && s == 0) {
      cout << "0 "
           << "0";
      t = 1;
    }
    if (t == 0) {
      if (m < s / 9 + 1 || s == 0) {
        cout << "-1 "
             << "-1";
      } else {
        small(m, s);
        great(m, s);
      }
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 small(long long m, long long s) {
  vector<long long> v;
  long long p = (s % 9 == 0) ? s / 9 - 1 : s / 9;
  for (long long i = 0; i < p; i++) {
    v.push_back(9);
    s -= 9;
  }
  if (m - p == 1) {
    v.push_back(s);
  }
  if (m - p >= 2) {
    s--;
    v.push_back(s);
    for (long long i = 0; i < m - p - 2; i++) {
      v.push_back(0);
    }
    v.push_back(1);
  }
  for (long long i = v.size() - 1; i >= 0; i--) {
    cout << v[i];
  }
  cout << " ";
}
void great(long long m, long long s) {
  vector<long long> v;
  long long k = s / 9;
  for (long long i = 0; i < k; i++) {
    v.push_back(9);
    s -= 9;
  }
  if (s > 0) {
    v.push_back(s);
    k++;
  }
  for (long long i = 0; i < m - k; i++) {
    v.push_back(0);
  }
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i];
  }
}
int main() {
  long long m, s, h;
  string k;
  vector<int> v;
  cin >> m >> s;
  if (s % 9 == 0) {
    long long t = 0;
    if (m == 1 && s == 0) {
      cout << "0 "
           << "0";
      t = 1;
    }
    if (t == 0) {
      if (m < s / 9 || s == 0) {
        cout << "-1 "
             << "-1";
      } else {
        small(m, s);
        great(m, s);
      }
    }
  } else if (s % 9 != 0) {
    long long t = 0;
    if (m == 1 && s == 0) {
      cout << "0 "
           << "0";
      t = 1;
    }
    if (t == 0) {
      if (m < s / 9 + 1 || s == 0) {
        cout << "-1 "
             << "-1";
      } else {
        small(m, s);
        great(m, s);
      }
    }
  }
  return 0;
}

```