codeforces875A Classroom Watch

A. Classroom Watch

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number n. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that n is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer x was given. The task was to add x to the sum of the digits of the number x written in decimal numeral system.

Since the number n on the board was small, Vova quickly guessed which x could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number n for all suitable values of x or determine that such x does not exist. Write such a program for Vova.

Input

The first line contains integer n (1 ≤ n ≤ 109).

Output

In the first line print one integer k — number of different values of x satisfying the condition.

In next k lines print these values in ascending order.

Examples

input

21

output

115

input

20

output

0

Note

In the first test case x = 15 there is only one variant: 15 + 1 + 5 = 21.

In the second test case there are no such x.

说下题意：给一个数n，让求的是x加上x的各个位之和等于n的数字的个数，并且输出x。

思路：假设x为len位数，x加上x的各个位之和与n的差一定不会超过len*9，所以暴力搜一遍就行啦。

#include <bits/stdc++.h>using namespace std;int n,a[10010];int main(){    while(~scanf("%d",&n)){        int temp = n, cnt = 0;        while(temp){            cnt ++; temp /= 10;        }        int ans = 0;        for(int i = n-cnt * 9; i <= n; i ++){            int j = i, sum = i;            while(j){                sum += (j % 10);                j /= 10;            }            if(sum == n) a[ans ++] = i;        }        printf("%d\n",ans);        for(int i = 0; i < ans; i ++){            if(!i) printf("%d",a[i]);            else printf(" %d",a[i]);        }        if(ans) printf("\n");    }    return 0;}