Problem:
Score : 600 points

Problem Statement

We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.

Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.

It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.

You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.

Constraints

0 ≤ K ≤ 50 \times 10^{16}

Input

Input is given from Standard Input in the following format:

K

Output

Print a solution in the following format:

N
a_1 a_2 ... a_N

Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.

Sample Input 1

0

Sample Output 1

4
3 3 3 3

Sample Input 2

1

Sample Output 2

3
1 0 3

Sample Input 3

2

Sample Output 3

2
2 2

The operation will be performed twice: [2, 2] -> [0, 3] -> [1, 1].

Sample Input 4

3

Sample Output 4

7
27 0 0 0 0 0 0

Sample Input 5

1234567894848

Sample Output 5

10
1000 193 256 777 0 1 1192 1234567891011 48 425

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}

//typedef
//------------------------------------------
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
typedef map<int,int> MII;
typedef queue<int> QI;
typedef queue<PII> QPII;
typedef stack<int> SI;
typedef stack<PII> SPII;
typedef deque<int> DI;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
typedef vector<bool> VB;
typedef vector<VB> VVB;
typedef vector<double> VD;
typedef vector<VD> VVD;
typedef vector<string> VS;
typedef vector<VS> VVS;
typedef vector<char> VC;
typedef vector<VC> VVC;
typedef vector<PII> VPII;
typedef vector<PLL> VPLL;
typedef priority_queue<int> PQGI; //大きい順
typedef priority_queue<int, VI, greater<int>> PQLI;
typedef priority_queue<PII> PQGP; 
typedef priority_queue<PII, VPII, greater<int>> PQLP;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define PB push_back
#define PF push_front
#define POB pop_back
#define POF pop_front
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define SORTR(c) sort((c).rbegin(), (c).rend())
#define LB lower_bound
#define UB upper_bound
#define NEXP next_permutation
#define FI first
#define SE second

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define FORR(i,a,b) for(int i = (b-1);i>=a;i--)
#define REPR(i,n) FORR(i,0,n)
#define BFOR(bit,a,b) for(int bit = (a); bit < (1<<(b)); ++bit)
#define BREP(bit,n) BFOR(bit,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int INF = INT_MAX/2;
const LL LINF = LLONG_MAX/3;
const int RINF = INT_MIN/2;
const LL RLINF = LLONG_MIN/3;
const LL MOD = 1e9+7;
const LL MODD = 998244353;
const int MAX = 510000;

inline bool Eq(double a, double b) { return fabs(b - a) < EPS; }

//other
//-------------------------------------------
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define COUT(x) cout << (x) << endl
#define COUT2(x,y) cout << (x) << " " << (y) << endl
#define COUT3(x,y,z) cout << (x) << " " << (y) << " " << (z) << endl
#define PR(x) cout << (x)
#define ENDL cout << endl
#define SPACE cout << " "
#define BC(x) __builtin_popcountll(x)

void cYes(){ COUT("Yes");exit(0); }
void cNo(){ COUT("No");exit(0); }
void cYES(){ COUT("YES");exit(0); }
void cNO(){ COUT("NO");exit(0); }
void cyes(){ COUT("yes");exit(0); }
void cno(){ COUT("no");exit(0); }
void cMINU(){ COUT(-1);exit(0); }
void cZERO(){ COUT(0);exit(0); }

VI dx = {1,0,-1,0,1,1,-1,-1};
VI dy = {0,1,0,-1,1,-1,1,-1};
VC dc = {'R','D','L','U'};
VS ds = {"R","D","L","U","RD","RU","LD","LU"};

LL Gcd(LL a, LL b) {return __gcd(a,b);}
LL Lcm(LL a, LL b) {return a/Gcd(a,b)*b;}

LL ModPow(LL A, LL N, LL M){
  LL res = 1;
  while(N>0){
    if(N&1) res = res * A % M;
    A = A * A % M;
    N >>= 1;
  }
  return res;
}

LL fac[MAX], finv[MAX], inv[MAX];

void ConInit(LL M) {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % M;
        inv[i] = M - inv[M%i] * (M / i) % M;
        finv[i] = finv[i - 1] * inv[i] % M;
    }
}

LL COM(int n, int k, LL M){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % M) % M;
}

template<class T> void VCout(vector<T> &v){
  int N = SZ(v);
  REP(i,N){
    cout << v.at(i);
    if(i == N-1) cout << endl;
    else cout << " ";
  }
  return;
}




int main() {
  
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(12);
  
  LL K; cin >> K;
  VLL a(50,49+K/50);
  
  K %= 50;
  
  REP(i,K){
    a.at(i) += 51;
    REP(j,50) a.at(i)--;
  }
  
  VCout(a);
  
  
  return 0;
}