第九周训练赛——C

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 xwas 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.

Example

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.

题意：

给你个数字，让你按规定拆分（看样里就能看懂）输出所有可能结果

思路：

采用for循环，注意一点就是如果n大于81，可以n减去81了（一开始将题目想复杂了）

代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>typedef long long ll;using namespace std;ll a[100];ll ans;ll n;bool pd(int x){int a=x,ans=x;while(a!=0){ans+=a%10;a/=10;}return ans==n;}int main(){ios::sync_with_stdio(false);while(cin>>n){int i;int s;if(n-81<1)s=1;else s=n-81;for(i=s;i<=n;i++)if(pd(i))a[ans++]=i;cout<<ans<<endl;for(int i=0;i<ans;i++)cout<<a[i]<<' ';cout<<endl;}return 0;}