（寒假CF） 比赛要

题意：给位数和位数和 求符合要求的最大值和最小值

//想法都有了，可是却没有做粗来（你484傻啊⁽⁽ ◟(눈_눈)◞ ⁾⁾）

Description

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 Input

Input

2 15

Output

69 96

Input

3 0

Output

-1 -1

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
    int m,s,s1,s2,i;
    int a[101],b[101];
    while(~scanf("%d %d",&m,&s))
    {
        int sum=0;
        if(m==1&&s==0)
        puts("0 0");
        else if(s>9*m||s==0)
        puts("-1 -1");
        else
        {
            s1=s2=s;
            for(i=m-1;i>=1;i--)
            {
                if(s1>9)
                {
                    a[i]=9;
                    s1-=9;
                }
                else if(s1>1)
                {
                    a[i]=s1-1;
                    s1=1;
                }
                else if(s1==1)
                    a[i]=0;
            }
            a[0]=s1;
            for(i=0;i<m;i++)
            {
                if(s2>9)
                {
                    b[i]=9;
                    s2-=9;
                }
                else if(s2>0)
                {
                    b[i]=s2;
                    s2=0;
                }
                else if(s2==0)
                    b[i]=0;    
            }
            for(i=0;i<m;i++)
            printf("%d",a[i]);
            printf(" ");
            for(i=0;i<m;i++)
            printf("%d",b[i]);
            printf("
");
        }
    }
    return 0;
}