Codeforces Round #277.5 (Div. 2)——C贪心—— Given Length and Sum of Digits

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

/*

   贪心

   对于min第一位先保证有1从后面开始加，如果加完之后还有s，说明NO，如果连1都没有那么NO

   对于max就从头往后如果s够就9不够就s

   其他特殊情况要判，比如1 0 坑点应该输出0 0。。其他带有0的都是-1 -1

*/

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

int main()

{

    int n,s;

    int a[110];

    while(~scanf("%d%d", &n, &s)){

        if(n == 1 && s == 0) {

            printf("0 0\n");

            continue;

        }

        if(n == 0 || s == 0 ){

            if( n== 1 && s == 0);

            else {

                printf("-1 -1\n");

                continue;

            }

        }

        memset(a, 0, sizeof(a));

        int temp = s;

        s --;

        a[1] = 1;

        int pos = n;

        int flag = 0;

        while(s > 0){

            if(s <= 8){

                a[pos--] += s;

                s = 0;

            }

            else{ a[pos--] = 9;

                s -= 9;

            }

            if(pos == 0) break;

        }

        int flag1 = 0;

        int sum1 = 0;

        for(int i = 1; i <= n; i++){

            sum1 += a[i];

            if(a[i] > 9)

                flag1 = 1;

        }

        if(sum1 != temp || flag1 == 1) printf("-1 ");

        else {

            for(int i = 1; i <= n ;i++)

                printf("%d", a[i]);

            printf(" ");

        }

        memset(a, 0, sizeof(a));

        s = temp;

        for(int i = 1; i <= n ;i++){

            if(s <= 8) {a[i] = s;s = 0;}

            else {a[i] = 9; s -= 9; }

        }

        int sum = 0;

        for(int i = 1; i <= n ; i++)

            sum += a[i];

        if(sum != temp || a[1] == 0) printf("-1\n");

        else {

            for(int i = 1; i <= n ;i++)

                printf("%d",a[i]);

            puts("");

        }

        

    }

        return 0;

}