/**

code by: zstu wxk

time: 2019/03/01

Problem Link: http://codeforces.com/contest/1130/problem/E

solve:

    如果构造的数列是 0 0 0 0 0 0 0 -1 + + + +(+代表大于0的数)

        那么 find_answer = sum(+) * len_+

        实际上的答案为 tot_len * [sum(+) - 1]

        即: sum(+) * len_+ + k = tot_len * [sum(+) - 1]

        接下来就是枚举tot_len就好了

**/

#include<bits/stdc++.h>

using namespace std;

#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);

#define LL long long

#define ULL unsigned LL

#define fi first

#define se second

#define pb push_back

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define lch(x) tr[x].son[0]

#define rch(x) tr[x].son[1]

#define max3(a,b,c) max(a,max(b,c))

#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;

const int inf = 0x3f3f3f3f;

const int _inf = 0xc0c0c0c0;

const LL INF = 0x3f3f3f3f3f3f3f3f;

const LL _INF = 0xc0c0c0c0c0c0c0c0;

const LL mod =  (int)1e9+;

const int N = 1e5 + ;

int n, k, t1, t2;

int a[N];

bool check(int x){

    for(int i = ; i * i <= x; ++i){

        if(x % i == ){

            t1 = i, t2 = x/i;

            t1 = n - t1;

            if(t1 < n -  && t2 < 1e6 * t1)

                return true;

        }

    }

    return false;

}

void Ac(){

    n = ;

    memset(a, , sizeof a);

    for(n = ; n >= ; --n){

        if(check(n+k)){

            for(int i = ; i <= t1; ++i){

                if(t2 >= 1e6){

                    a[i] = 1e6;

                    t2 -= 1e6;

                }

                else{

                    a[i] = t2;

                    t2 = ;

                }

            }

            a[t1+] = -;

            printf("%d\n", n);

            reverse(a+, a++n);

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

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

            }

            break;

        }

    }

}

int main(){

    while(~scanf("%d", &k)){

        Ac();

    }

    return ;

}