C. Given Length and Sum of Digits...
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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).

Sample test(s)
input
2 15
output
69 96
input
3 0
output
-1 -1


#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;

bool can(int m, int s)
{
    if(s >= 0 && 9*m >= s) return true;
    else return false;
}
int main()
{
    int m,s;
    cin>>m>>s;
    if(!can(m,s))
    {
        cout<<"-1"<<" "<<"-1"<<endl;
        return 0;
    }
    if(m == 1)
    {
        if(s >= 10)
        {
            cout<<"-1"<<" "<<"-1"<<endl;
        }
        else cout<<s<<" "<<s<<endl;
    }
    else {
        if(s == 0) cout<<"-1"<<" "<<"-1"<<endl;
        else {

            string minn, maxn;
            int sum = s;

            for(int i = 1; i <= m; i++)
                for(int j = 0; j < 10; j++)
            {
                if((j > 0 || (j == 0 && i > 1) ) && can(m - i, sum - j))
                   {
                       minn += char('0' + j);
                       sum -= j;
                       break;
                   }
            }



            sum = s;
               for(int i = 1; i <= m; i++)
                for(int j = 9; j >= 0; j--)
            {
                if(can(m - i, sum - j))
                   {
                       maxn += char('0' + j);
                       sum -= j;
                       break;
                   }
            }

            cout<<minn<<" "<<maxn<<endl;

        }
    }
    return 0;
}