codeforces 489C Given Length and Sum of Digits...

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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).

Examples

input

2 15

output

69 96

input

3 0

output

-1 -1

用数组，找最小值比较难，最大值很简单，最小值的找法是第一位先赋值1，然后从后往前能赋9的赋9，不能的把剩下的全给它（在0~9），如果还有多余的给第一位。

注意几组数据：输入1和0，应输出0和0；m=0，直接就是-1和-1；m>1并且s=0，输出-1和-1，其余的就比较正常了，具体的看代码！

#include<stdio.h>
#include<string.h>
#include<string>
using namespace std;
int main()
{
    int a[110],b[110];
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    int m,s;
    scanf("%d%d",&m,&s);
    if(m==0)
    {
        printf("-1 -1\n");
        return 0;
    }
    if(m==1&&s==0)
    {
        printf("0 0\n");
        return 0;
    }
    if(m>1&&s==0)
    {
        printf("-1 -1\n");
        return 0;
    }
    int t=s-1;
    a[0]=1;
    for(int i=m-1;i>=1;i--)
    {
        if(t>=9)
        {
            a[i]=9;
            t-=9;
        }
        else
        {
            a[i]=t;t=0;
            break;
        }
    }
    if(t>8)
    {
       printf("-1 -1\n");
       return 0;
    }
    if(t>0)
        a[0]=t+a[0];
    t=s;
    for(int i=0;i<m;i++)
    {
        if(t>=9)
        {
            b[i]=9;t-=9;
        }
        else
        {
            b[i]=t;
            break;
        }
    }
    for(int i=0;i<m;i++)
        printf("%d",a[i]);
    printf(" ");
    for(int i=0;i<m;i++)
        printf("%d",b[i]);
    printf("\n");
    return 0;
}