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)

// AtCoder.cpp : このファイルには 'main' 関数が含まれています。プログラム実行の開始と終了がそこで行われます。
//

// AtCoder.cpp : このファイルには 'main' 関数が含まれています。プログラム実行の開始と終了がそこで行われます。
//


#define _USE_MATH_DEFINES
#include<math.h>

//#include<cmath>

#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>



//#include "Ants.h"
using namespace std;
typedef long long ll;
#define rad_to_deg(rad) (((rad)/2/M_PI)*360)
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define rep(i,n) for(int i=0;i<n;i++)
#define show(s) cout<<s<<endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define LINF (1000000000000000000ll)
#define MOD (1e9+7)
#define rep1(i,n) for(int i =1;i<=n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;--i)
typedef pair < ll, ll> P;
ll gcd(ll a, ll b) {//最大公約数を求める///最大公倍数a*b/gcd(a,b)
	if (b == 0) return a;
	return gcd(b, a % b);
}
ll lcd(ll a, ll b) {//最小公倍数を求める　b=0のときaを返す
	if (b == 0) return a;
	return (a / gcd(a, b)) * b;
}
vector<int> divnum(ll num) {
	int dig;
	vector<int>p;
	while (num) {
		dig = num % 10;
		p.push_back(dig);
		num /= 10;
	}
	return p;
}
int digiter(ll num) {
	int dig;
	vector<int>p;
	while (num) {
		dig = num % 10;
		p.push_back(dig);
		num /= 10;
	}
	return p.size();
}
vector<int> convertstring(string s) {
	vector<int> d;
	ll n = s.size();
	rep(i, n) {
		d.push_back(s[i] - '0');
	}
	return d;
}
map< ll, int > prime_factor(ll n) {
	map< ll, int > ret;
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret[i]++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}
vector<ll> divisor(ll n) {
	vector<ll>res;
	for (ll i = 1; i * i <= n; i++) {
		if (n % i != 0)continue;
		res.push_back(i);
		if (i * i != n)res.push_back(n / i);
	}
	return res;
}


ll myPow(ll x, ll n, ll m) {//x^n
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return myPow(x * x % m, n / 2, m);
	else
		return x * myPow(x, n - 1, m) % m;
}
std::vector<int> Eratosthenes(const int N)//素数列挙
{
	std::vector<bool> is_prime(N + 1);
	for (int i = 0; i <= N; i++)
	{
		is_prime[i] = true;
	}
	std::vector<int> P;
	for (int i = 2; i <= N; i++)
	{
		if (is_prime[i])
		{
			for (int j = 2 * i; j <= N; j += i)
			{
				is_prime[j] = false;
			}
			P.emplace_back(i);
		}
	}
	return P;
}
const ll mod = 998244353;

struct mint {

	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x% mod + mod) % mod) {}
	
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; }
	mint operator+(const mint a) const { return mint(*this) += a; }
	mint operator-(const mint a) const { return mint(*this) -= a; }
	mint operator*(const mint a) const { return mint(*this) *= a; }
	
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const { return pow(mod - 2); }
	mint& operator/=(const mint a) { return *this *= a.inv(); }
	mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
struct combination {//comb(a,b)でよぶaCb

	vector<mint> fact, ifact;
	combination(ll n) :fact(n + 1), ifact(n + 1) {
		assert(n < mod);
		fact[0] = 1;
		for (ll i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (ll i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(ll n, ll k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
}comb(1000005);//comb貯め
mint f(ll n, ll k) {
	if (n < 0) return 0;
	// nPk = nCk * k!
	mint res = comb(n, k);
	res *= comb.fact[k];
	return res;
}

/*int Q;


struct SegmentTree {
private:
	int n;
	vector<int> node;

public:
	SegmentTree(vector<int> v) {//SegmentTree seg(vector<int> v(N,INF))で呼び出し 
		int sz = v.size();
		n = 1; while (n < sz) n *= 2;
		node.resize(2 * n - 1, INF);
		for (int i = 0; i < sz; i++) node[i + n - 1] = v[i];
		for (int i = n - 2; i >= 0; i--) node[i] = min(node[2 * i + 1], node[2 * i + 2]);
	}

	void update(int x, int val) {
		x += (n - 1);
		node[x] = val;
		while (x > 0) {
			x = (x - 1) / 2;
			node[x] = min(node[2 * x + 1], node[2 * x + 2]);
		}
	}

	int getmin(int a, int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		if (r <= a || b <= l) return INF;
		if (a <= l && r <= b) return node[k];

		int vl = getmin(a, b, 2 * k + 1, l, (l + r) / 2);
		int vr = getmin(a, b, 2 * k + 2, (l + r) / 2, r);
		return min(vl, vr);
	}
};
// 1-indexedなので注意。
struct BIT {
private:
	vector<int> bit;
	int N;
public:
	BIT(int size) {
		N = size;
		bit.resize(N + 1);
	}

	// 一点更新です
	void add(int a, int w) {//a番目をwに更新
		for (int x = a; x <= N; x += x & -x) bit[x] += w;
	}

	// 1~Nまでの和を求める。
	int sum(int a) {
		int ret = 0;
		for (int x = a; x > 0; x -= x & -x) ret += bit[x];
		return ret;
	}
};*/

ll K;
int main() {
	cin >> K;
	const ll d = K / 50LL, t = K % 50LL;
	const ll x = 49;
	rep(i, t) {
		ll a = 51 - t;
		cout << a+x <<" ";
	}
	rep(i, 50 - t) {
		ll a = -t;
		cout << a+x <<" ";
	}
	cout << endl;
}