Codeforces Round #277.5 (Div. 2) (question C)

Codeforces Round #277.5 (Div. 2) (question C)
C. Given Length and Sum of Digits... time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

You have a positive integerMAnd a non-negative integerS. Your task is to find the smallest and the largest of the numbers that have lengthMAnd sum of digitsS. The required numbers shocould be non-negative integers written in the decimal base without leading zeroes.

Input

The single line of the input contains a pair of integersM,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 
 
  #include #include 
  
   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"<
   
    = 10)        {            cout<<"-1"<<" "<<"-1"<
    
      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<