（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

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int m,s,s1,s2,i;
 8     int a[101],b[101];
 9     while(~scanf("%d %d",&m,&s))
10     {
11         int sum=0;
12         if(m==1&&s==0)
13         puts("0 0");
14         else if(s>9*m||s==0)
15         puts("-1 -1");
16         else
17         {
18             s1=s2=s;
19             for(i=m-1;i>=1;i--)
20             {
21                 if(s1>9)
22                 {
23                     a[i]=9;
24                     s1-=9;
25                 }
26                 else if(s1>1)
27                 {
28                     a[i]=s1-1;
29                     s1=1;
30                 }
31                 else if(s1==1)
32                     a[i]=0;
33             }
34             a[0]=s1;
35             for(i=0;i<m;i++)
36             {
37                 if(s2>9)
38                 {
39                     b[i]=9;
40                     s2-=9;
41                 }
42                 else if(s2>0)
43                 {
44                     b[i]=s2;
45                     s2=0;
46                 }
47                 else if(s2==0)
48                     b[i]=0;    
49             }
50             for(i=0;i<m;i++)
51             printf("%d",a[i]);
52             printf(" ");
53             for(i=0;i<m;i++)
54             printf("%d",b[i]);
55             printf("
");
56         }
57     }
58     return 0;
59 }

（寒假CF） 比赛要

（寒假CF）比赛要