Discription

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

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

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

21

1
15

20

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.

很顯然每位數的和的貢獻不可能太大，對於題目的範圍來說肯定是<100的，所以枚舉比n小100之內的數就行了。。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int num[1005];
int n,m,k;

inline int get(int x){
	int an=0,nxt;
	while(x){
		nxt=x/10;
		an+=x-nxt*10;
		x=nxt;
	}
	return an;
}

int main(){
	cin>>n;
	k=max(1,n-100);
	for(int i=k;i<n;i++) if(i==n-get(i)) num[++m]=i;
	
	printf("%d\n",m);
	for(int i=1;i<=m;i++) printf("%d\n",num[i]);
	return 0;
}

「Codeforces Round #441」 Classroom Watch