CodeForces 875A Classroom Watch

Description

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.

Example

Input

21

Output

1
15

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.

题目大意：

给你一个数n， 请找到一个数x， 使得x加上自己本身每一位的和等于n， 输出符合条件的x的个数。

解题思路：

首先能想到的是对于1e9的数据量， 暴力解决问题必定会超时， 那么我们就想到对于x， x的大小一定不会小于n减去n的每位之和， 因为如果x比当前这个数还要小的话根本不可能构成这个数字， 这样就不用每次都从1跑到n地去找符合的数字个数了。 

代码：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

using namespace std;

/* 

ios::sync_with_stdio(false);

*/

typedef long long ll;
typedef unsigned long long ull;
const int dir[5][2] = {0, 1, 0, -1, 1, 0, -1, 0, 0, 0};
const int inf = 0x7fffffff;
const int mod = 1000000;
const int Max = 1000000;

int temp[10], ans[Max];

void check(char *s) {
    int index = 0, dec, sum = 0, cnt = 0;
    dec = atoi(s);
    memset(temp, 0, sizeof(temp));
    for (int i = 0; i < strlen(s); ++i)
        sum += s[i];
    for (int i = max((dec - sum), 0); i <= dec; ++i) {
        int temp = i, t = i;
        while (temp) {
            t += (temp % 10);
            temp /= 10;
        }
        if (t == dec) {
            ans[cnt++] = i;
        }
    }
    printf("%d\n", cnt);
    for (int i = 0; i < cnt; ++i) {
        printf("%d%c", ans[i], i == cnt - 1 ? '\n' : ' ');
    }
}

int main() {
    char num[10];
    scanf("%s", num);
    check(num);
    return 0;
}